3.1 需要衍生因素:曲线的切线

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  需要衍生因素:曲线的切线

  播放段落

  Kevin is learning about the basis of calculus and what calculus is actually used for. Unfortunately, Kevin does not understand why calculus is sometimes necessary to find the equation of a line. In Algebra 1, he learned you can find the equation of a line if you are given two points. You find the slope of the line by dividing the up/down difference in the points by the left/right difference, then you use one of the points and the slope to find the y-intercept.
  ::Kevin正在学习微积分的基础以及实际使用的微积分是什么。 不幸的是, Kevin不明白为什么微积分有时对于找到线的方程是必要的。 在代数 1 中, 他了解到如果给定两个点, 您可以找到线的方程 。 您可以通过将点的上下差与左/ 右差进行分隔, 找到线的斜度, 然后用一个点和斜度来找到 y 界面 。

  播放段落

  Kevin's teacher, Mr. Banner, offered him extra credit if he could find the slope of a line for the points (4,5) and (4,5) using the method he learned in Algebra 1. Can you see what Mr. Banner did? What is Kevin going to find as he works on those problems?
  ::Kevin的老师Banner先生 利用他在代数1中学到的方法 给他额外贷款 如果他能找到点数(4,5)和点数(4,5)的斜坡

  播放段落

  Tangents to a Curve
  ::曲线的切线

  播放段落

  Recall from algebra, if points P ( x ₀ , y ₀ ) and Q ( x ₁ , y ₁ ) are two different points on the curve y = f ( x ), then the slope of the secant line connecting the two points is given by
  ::从代数中召回的代数,如果点P(x0,y0)和点Q(x1,y1)是曲线y = f(x)两个不同的点,则连接两个点的松动线的斜度由

  	$\begin{align*}m_{sec}\end{align*}$	$\begin{align*}= \frac {y_1-y_0}{x_1-x_0} = \frac {f(x_1)-f(x_0)}{x_1-x_0}\end{align*}$	(1)

  

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  Of course, if we let the point x ₁ approach x _o then Q will approach P along the graph f and thus the slope of the secant line will gradually approach the slope of the tangent line as x ₁ approaches x ₀ . Therefore, (1) becomes
  ::当然,如果我们让点 x1 接近x1 方向xo, Q将沿着图f 接近P, 分离线的斜度将随着 x1 接近x0 逐渐接近正切线的斜度。 因此, (1) 成为

  		$\begin{align*}m_{sec}\end{align*}$	$\begin{align*}= \lim_{x_1 \to x_0} \frac {f(x_1)-f(x_0)}{x_1-x_0} \end{align*}$	(2)

  播放段落

  To simplify our notation, if we let h = x ₁ − x ₀ , then x ₁ = x ₀ + h and x ₁ → x ₀ becomes equivalent to h → 0. This means that (2) becomes
  ::为了简化我们的标记,如果我们让 h = x1 -x0,那么如果让 h = x1 -x0,那么x1 = x0 + h + 和 x1 + h + 和 x1 x1 x0 等于 h + 0。 这意味着 (2) 变成

  		$\begin{align*}m_{sec}\end{align*}$	$\begin{align*}= \lim_{h \to 0} \frac {f(x_0+h)-f(x_0)}{h} \end{align*}$

  播放段落 The Slope of a Tangent Line ::延时线的斜线 If the point P ( x 0 , y 0 ) is on the curve f , then the tangent line at P has a slope that is given by $\begin{align*}m_{tan}=\lim_{h \to 0} \frac {f(x_0+h)-f(x_0)}{h} \end{align*}$ 播放段落 provided that the limit exist. ::前提是存在限额。

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  Recall that the equation of the tangent line through point ( x ₀ , y ₀ ) with slope m is the point-slope form of a line: y − y ₀ = m _tan ( x − x ₀ ).
  ::回顾通过点(x0, y0)与斜度米的正切线的方程式是线的点斜形:y-y0 = mtan(x- x0) = mtan(x- x0)。

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were given a problem about Kevin, who is having trouble understanding calculus.
  ::早些时候,你被给了凯文一个问题, 凯文很难理解微积分。

  播放段落

  Mr. Banner asked Kevin to find the equation of a line given the points (4,5) and (4,5). The points (4, 5) and (4, 5) are the same, so the $\begin{align*}\frac{rise}{run}\end{align*}$ would be $\begin{align*}\frac{0}{0}\end{align*}$ - Kevin was just introduced to the need for differential calculus!
  ::Banner先生要求Kevin根据点数(4,5)和点数(4,5)找到一条线的等式。 点数(4,5)和点数(4,5)相同,所以涨幅将是00,Kevin刚刚被介绍到需要差别微积分!

  播放段落

  Example 2
  ::例2

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  Find line tangent to the curve f ( x ) = x ³ that passes through point P (2,8).
  ::查找通过点P(2,8)的曲线 f(x) = x3 的正切线。

  播放段落

  Since P ( x ₀ , y ₀ ) = (2, 8), using the slope of the tangent equation we have
  ::自P(x0, y0) = (2, 8) 以来, 使用正切方程的斜度

  		$\begin{align*}m_{tan}\end{align*}$	$\begin{align*}= \lim_{h \to 0} \frac {f(x_0+h)-f(x_0)}{h} \end{align*}$
  and we get
  		$\begin{align*}m_{tan}\end{align*}$	$\begin{align*}= \lim_{h \to 0} \frac {f(2+h)-f(2)}{h} \end{align*}$
  			$\begin{align*}= \lim_{h \to 0} \frac {(h^3+6h^2+12h+8)-8}{h} \end{align*}$
  			$\begin{align*}= \lim_{h \to 0} \frac {h^3+6h^2+12h}{h} \end{align*}$
  			$\begin{align*}= \lim_{h \to 0} (h^2+6h+12) \end{align*}$
  			$\begin{align*}= 12\end{align*}$

  播放段落

  Thus the slope of the tangent line is 12. Using the point-slope formula above, we find that the equation of the tangent line is y - 8 = 12 ( x - 2) or y = 12 x - 16.
  ::因此,正切线的斜坡是12。 使用上面的点倾斜公式,我们发现正切线的方程式是y - 8 = 12 (x - 2) 或y = 12x - 16。

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  Example 3
  ::例3

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  If f ( x ) = x ² − 3,find f' ( x ) and use the result to find the slope of the tangent line at x = 2 and x = −1.
  ::如果f(x) = x2 - 3, ff' (x) 并使用结果在 x = 2 和 x = = 1 时找到正切线的斜度,则f(x) = x2 = 3, ff' (x) 。

  播放段落

  Since $\begin{align*}f'(x)= \lim_{h \to 0} \frac {f(x+h)-f(x)}{h} \end{align*}$ then
  ::从f_(x) =limh_0f(x+h) -f(x) 到(x) h

  		$\begin{align*}f'(x)\end{align*}$	$\begin{align*}= \lim_{h \to 0} \frac {\left [ {(x+h)^2-3} \right ]- \left [ {x^2-3} \right ]}{h} \end{align*}$
  			$\begin{align*}= \lim_{h \to 0}\frac {x^2+2xh+h^2-3-x^2+3}{h}\end{align*}$
  			$\begin{align*}= \lim_{h \to 0} \frac {2xh+h^2}{h}\end{align*}$
  			$\begin{align*}= \lim_{h \to 0} (2x+h)\end{align*}$
  			$\begin{align*}= 2x\end{align*}$

  播放段落

  To find the slope, we simply substitute x = 2 into the result f' ( x ):
  ::为了找到斜坡,我们只需将x=2替换为结果f'(x):

  		f '( x )	= 2 x
  		f'(2)	=2(2)
  			= 4
  and
  		f'(x)	=2x
  		f'(-1)	=2(-1)
  			= -2

  播放段落

  Thus slope of the tangent line at x = 2 and x = −1 are 4 and −2 respectively.
  ::因此,x = 2和x =-1的正切线斜坡分别为4和-2。

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  Example 4
  ::例4

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  Find the slope of the tangent line to the curve y = 1/ x that passes through the point (1, 1).
  ::查找穿过点(1,1)的曲线 y = 1/x 的正切线斜度。

  播放段落

  Using the slope of the tangent formula,
  ::使用正切公式的斜坡,

  		$\begin{align*}f'(x)\end{align*}$	$\begin{align*}= \lim_{h \to 0} \frac {f(x+h)-f(x)}{h}\end{align*}$
  and substituting $\begin{align*}y=\frac{1}{x}\end{align*}$
  		$\begin{align*}f'(x)\end{align*}$	$\begin{align*}= \lim_{h \to 0} \frac {\left ( \frac{1}{x+h} \right )-\frac {1}{x}}{h}\end{align*}$
  			$\begin{align*}= \lim_{h \to 0} \frac {\frac {x-x-h}{x \left ({x+h} \right)}}{h}\end{align*}$
  			$\begin{align*}= \lim_{h \to 0} \frac {x-x-h}{hx \left ({x+h} \right)}\end{align*}$
  			$\begin{align*}= \lim_{h \to 0} \frac {-h}{hx \left ({x+h} \right)}\end{align*}$
  			$\begin{align*}= \lim_{h \to 0} \frac {-1}{x \left ({x+h} \right)}\end{align*}$
  			$\begin{align*}= \frac {-1}{x^2}\end{align*}$
  For x = 1, the slope is
  		$\begin{align*}f'(x)\end{align*}$	$\begin{align*}= \frac {-1}{1} = -1\end{align*}$
  			$\begin{align*}= -1\end{align*}$

  播放段落

  Thus the slope of the tangent line at x = 1 for the curve y = 1/ x is m = −1. To find the equation of the tangent line, we simply use the point-slope formula,
  ::因此,在 x = 1 的正切线的斜度为 y = 1/x 的曲线 y = 1/x = m = = 1 。 要找到正切线的方程, 我们只需使用点倾斜公式,

  		y - y 0	= m ( x - x 0 )
  Where ( x 0 , y 0 ) = (1, 1).
  		y - 1	= -1( x - 1)
  		y	= - x + 1 + 1
  		y	= - x + 2

  播放段落

  So the equation of the tangent line is y = - x + 2.
  ::因此,正切线的方程式是 y = -x + 2。

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  Example 5
  ::例5

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  Given the function $\begin{align*}y = \frac{1}{2} x^2\end{align*}$ and the values of $\begin{align*}x_0 = 3\end{align*}$ and $\begin{align*}x_1 = 4\end{align*}$ , find:
  ::根据 y= 12x2 函数以及 x0= 3 和 x1= 4 的值,找到 :

  播放段落
  1. The average rate of change of y with respect to x over the interval [ x ₀ , x ₁ ].
     ::x在间隔[x0, x1]内的平均y变化率。

  播放段落

  Identify the two points by substituting 3 and 4 in for x in the function $\begin{align*}f(x) = \frac{1}{2}x^2\end{align*}$
  ::在函数 f(x) = 12x2 中以 x 替换 3 和 4 来识别两个点

  播放段落

  Substitute the two points (3, 4.5) and (4, 8) into the average rate of change formula: $\begin{align*}m = \frac{y_1 - y_0}{x_1 - x_0}\end{align*}$
  ::将两个点(3,4.5)和两个点(4,8)替换为平均变化率公式: m=y1-y0x1-x0

  播放段落

  Average rate of change = $\begin{align*}\frac{7} {2}\end{align*}$
  ::平均变动率=72

  播放段落
  2. The slope of the secant line connecting x ₀ and x _1.
     ::连接 x0 和 x1 的松散线的斜坡。

  播放段落

  The slope of the secant line between $\begin{align*}x_0\end{align*}$ and $\begin{align*}x_1\end{align*}$ is the slope between $\begin{align*}(3, 4.5)\end{align*}$ and $\begin{align*}(4, 8)\end{align*}$ , which is $\begin{align*}\frac{7} {2}\end{align*}$ .
  ::x0和x1之间的分离线的斜坡是(34.5)和(4.8)之间的斜坡,即72。

  播放段落
  3. The instantaneous rate of change of y with respect to x at x ₀ .
     ::xxxxxxx0的y的瞬时变化率。

  播放段落

  Instantaneous rate of change is the slope at x = 3.
  ::瞬间变化速度是x=3的斜坡。

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  Use the formula: $\begin{align*}\frac{f(x + h) - f(x)}{h}\end{align*}$ where $\begin{align*}f(x) = \frac{1}{2}x^2\end{align*}$ and $\begin{align*}x = 3\end{align*}$
  ::使用公式 : f( x+h) - f( x) h 的 f( x) = 12x2 和 x= 3

  播放段落

  $\begin{align*}\frac{f(3 + h) - f(3)}{h}\end{align*}$ ..... Substitute 3 for x
  ::f( 3+h) - f(3)h. x 替代 3

  播放段落

  $\begin{align*}\frac{\frac{1}{2}(3 + h)^2 - \frac{1}{2}(3)^2}{h}\end{align*}$ ..... Replace $\begin{align*}f(x) \to \frac{1}{2}x^2\end{align*}$
  ::12( 3+h) 2- 12(3)2h. 替换 f( x) 12x2

  播放段落

  $\begin{align*}\frac{\frac{1}{2}(9) + \frac{1}{2}(6h) + \frac{1}{2}h^2 - \frac{1}{2}9}{h}\end{align*}$ ..... FOIL and Distribute the 1/2
  ::12(9)+1212(6h)+12h2-129h.FOIL和分配1/2

  播放段落

  $\begin{align*}\frac{6h + h^2}{2h}\end{align*}$ ..... Simplify
  ::6h+h22h.... 简化

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  $\begin{align*}3+\frac{h}{2}\end{align*}$ ..... Simplify again
  ::3+h2.... 再次简化

  播放段落

  $\begin{align*}3\end{align*}$ ..... As $\begin{align*}h \to 0\end{align*}$
  ::. . . . . . . . . 成为 h0

  播放段落

  $\begin{align*}\therefore\end{align*}$ the instantaneous slope at x = 3 is 3
  ::X = 3 的瞬时斜度为 3

  播放段落
  4. The slope of the tangent line at x ₁ .
     ::x1 相切线的斜坡。

  播放段落

  The slope of the tangent at 4 is the same as the instantaneous rate of change at $\begin{align*}x = 4\end{align*}$
  ::4时相切度的斜坡与x=4时的瞬时变化率相同

  播放段落

  This is the same series of steps as with x = 3 above
  ::这是与以上 x = 3 相同的一系列步骤

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  $\begin{align*}\therefore\end{align*}$ the slope at x = 4 is 4
  ::X = 4 的斜坡是 4

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  Example 6
  ::例6

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  Given the function $\begin{align*}f(x) = \frac{1}{x}\end{align*}$ and the values $\begin{align*}x_{0} = 2\end{align*}$ and $\begin{align*}x_1 = 3\end{align*}$ , find:
  ::根据函数 f( x) = 1x 和值 x0= 2 和 x1= 3, 查找 :

  播放段落
  1. The average rate of change of y with respect to x over the interval [ x ₀ , x ₁ ].
     ::x在间隔[x0, x1]内的平均y变化率。

  播放段落

  Identify the two points by substituting 2 and 3 in for x in the function $\begin{align*}f(x) = \frac{1}{x}\end{align*}$ to get $\begin{align*}(2, \frac{1}{2}) | (3, \frac{1}{3})\end{align*}$
  ::在函数 f(x) = 1x 中以 x 替换 2 和 3 来识别两个点( 2, 12) ( 3, 13) 。

  播放段落

  Substitute the two points $\begin{align*}(2, \frac{1}{2}) | (3, \frac{1}{3})\end{align*}$ into the average rate of change formula: $\begin{align*}m = \frac{y_1 - y_0}{x_1 - x_0}\end{align*}$
  ::将两个点(2,12)(3,13)替代为平均变化率公式: m=y1-y0x1-x0

  播放段落

  Average rate of change = $\begin{align*}\frac{-1}{6}\end{align*}$
  ::平均变化率=-16

  播放段落
  2. The slope of the secant line connecting x ₀ and x ₁ .
     ::连接 x0 和 x1 的松散线的斜坡。

  播放段落

  The slope of the secant line between $\begin{align*}x_0\end{align*}$ and $\begin{align*}x_1\end{align*}$ is the slope between $\begin{align*}(2, \frac{1}{2})\end{align*}$ and $\begin{align*}(3, \frac{1}{3})\end{align*}$ , which is $\begin{align*}\frac{-1}{6}\end{align*}$ .
  ::x0和x1之间的分离线的斜坡是(2,12)和(3,13)之间的斜坡,即-16。

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  3. The instantaneous rate of change of y with respect to x at x ₀ .
     ::xxxxxxx0的y的瞬时变化率。

  播放段落

  Instantaneous rate of change at x ₀ is the slope at x = 2.
  ::x0时的瞬时变化速度是斜面x=2。

  播放段落

  Use the formula: $\begin{align*}\frac{f(x + h) - f(x)}{h}\end{align*}$ where $\begin{align*}f(x) = \frac{1}{x}\end{align*}$ and $\begin{align*}x = 2\end{align*}$
  ::使用公式 : f( x+h) - f( x) h, 其中 f( x)=1x 和 x=2

  播放段落

  $\begin{align*}\frac{f(2 + h) - f(2)}{h}\end{align*}$ ..... Substitute 2 for x
  ::f(2+h) - f(2)h. x 替代 2

  播放段落

  $\begin{align*}\frac{\frac{1}{2 + h} - \frac{1}{2}}{h}\end{align*}$ ..... Replace $\begin{align*}f(x) \to \frac{1}{x}\end{align*}$
  ::12+h- 12h.... 替换 f(x)1x

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  $\begin{align*}(\frac{1}{2 + h} - \frac{1}{2}) \cdot \frac{1}{h}\end{align*}$ ..... We had a fraction divided by a fraction, invert to multiply
  :12+h-12)1h.。我们有一个小数除以一个小数,反转成乘法

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  $\begin{align*}\frac{(2)(1)}{2(2 + h)} - \frac{(2 + h)(1)}{2(2 + h)} \cdot \frac{1}{h}\end{align*}$ ..... Set common denominators
  :2)(2)2(2+h)-(2+h)(1)(2)(2)(2+h)-(2+h)-1h.。

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  %7D"> $\begin{align*}\frac{(2) - (2 + h)}{(2 + h)(2)(h)}\end{align*}$ ..... Simplify
  :2)-2+(2+h)(2+h)(2+h)(2)(2)......简化

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  $\begin{align*}\frac{-h}{4h +2h^2}\end{align*}$ ..... Simplify again
  ::-h4h+2h2...。再次简化

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  $\begin{align*}\frac{-1}{4 + 2h}\end{align*}$ ..... once more (canceling the h )
  ::- 14+2h... ... . 再来一次(取消h)

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  $\begin{align*}\frac{-1}{4}\end{align*}$ ..... As $\begin{align*}h \to 0\end{align*}$
  ::-14... ... . 成为h*0

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  $\begin{align*}\therefore\end{align*}$ the instantaneous slope at x = 2 is $\begin{align*}\frac{-1} {4}\end{align*}$
  ::X = 2 的瞬时斜度为 ~ 14

  播放段落
  4. The slope of the tangent line at x ₁ .
     ::x1 相切线的斜坡。

  播放段落

  The slope of the tangent at 3 is the same as the instantaneous rate of change at $\begin{align*}x = 3\end{align*}$
  ::3时相切度的坡度与x=3时的瞬时变化率相同。

  播放段落

  This is the same series of steps as with x = 2 above
  ::这是与以上 x = 2 相同的一系列步骤

  播放段落

  $\begin{align*}\therefore\end{align*}$ the slope at x = 3 is $\begin{align*}\frac{-1} {9}\end{align*}$
  ::x = 3 的斜坡是 ~ 19

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  Review
  ::回顾

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  1. What is the line connecting two points $\begin{align*}(x_0, y_0)\end{align*}$ and $\begin{align*}(x_1, y_1)\end{align*}$ on a curve called?
     ::在所谓的曲线上连接两个点(x0,y0)和(x1,y1)的线是什么?
  播放段落
  2. As $\begin{align*}(x_0, y_0)\end{align*}$ gets immeasurably close to $\begin{align*}(x_1, y_1)\end{align*}$ the term describing the line between them becomes: "the ____________ line"
     ::由于 (x0, y0) 接近 (x1,y1) 的程度不可估量, 描述它们之间的线的术语变成 : “ 线 ”
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  3. The expression $\begin{align*}f(x_0 + h) - f(x_0)\end{align*}$ is used to describe what distance in the process of finding the slope of a tangent line?
     ::表达式 f( x0+h) - f( x0) 用于描述查找正切线斜坡过程中的距离 ?
  播放段落
  4. When calculating the slope of a tangent, what value is assumed to go to 0 as the two chosen points get closer and closer?
     ::当计算正切的斜坡时,当两个选定点越来越接近时,假定值为0是多少?
  播放段落
  5. What does the concept of limit, discussed in prior lessons, have to do with finding the slope of a line tangent to a curve?
     ::以往经验教训中讨论的限制概念与找到曲线正切线的斜坡有何关系?

  播放段落

  Find the equation of the tangent line:
  ::查找正切线的方程式 :

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  6. What is the equation of the tangent line at $\begin{align*}x = -3\end{align*}$ assuming that $\begin{align*}r(-3) = - 5\end{align*}$ and $\begin{align*}r'(-3) = 1\end{align*}$ ?
     ::x% 3 上的正切线的等式是什么? 假设 r(- 3) 5 和 r(- 3) = 1 ?
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  7. What is the equation of the tangent line at $\begin{align*} x = 1\end{align*}$ assuming that $\begin{align*}r(1) = 3\end{align*}$ and $\begin{align*}r'(1) = -5\end{align*}$ ?
     ::x=1时的正切线的方程式是多少? 假设r(1)=3和r(1)=5?
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  8. What is the equation of the tangent line at $\begin{align*}x = 2\end{align*}$ assuming that $\begin{align*}g(2) = 1\end{align*}$ and $\begin{align*}g'(2) = -3\end{align*}$ ?
     ::x=2 上的正切线的方程式是多少? 假设 g(2)=1 和 g(2)=3 ?
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  9. What is the equation of the tangent line at $\begin{align*}x = 4\end{align*}$ assuming that $\begin{align*}u(4) = 4\end{align*}$ and $\begin{align*}u'(4) = 3\end{align*}$ ?
     ::x=4时的正切线的方程式是多少,假设是 u(4)=4 和 u(4)=3?
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  10. What is the equation of the tangent line at $\begin{align*}x = -4\end{align*}$ assuming that $\begin{align*}t(-4) = 2\end{align*}$ and $\begin{align*}t'(-4) = 5\end{align*}$ ?
      ::x4的正切线的方程式是多少? 假设 t(-4)=2和 t(-4)=5?

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  Find the equation of the tangent line:
  ::查找正切线的方程式 :

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  11. Find the equation of the tangent line to the graph of $\begin{align*}h(x) = -5x^3 - 3x^2 + x + 3 \end{align*}$ at $\begin{align*}x = 1\end{align*}$
      ::查找 x=1 时 h( x) =5x3- 3x2+x+3 图形的正切线的方程
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  12. Find the equation of the tangent line to the graph of $\begin{align*}t(x) = -2x\end{align*}$ at $\begin{align*}x = -2\end{align*}$
      ::查找 x2 上的 t( x) % 2x 图形的正切线方程式的方程式。
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  13. Find the equation of the tangent line to the graph of $\begin{align*}m(x) = 3x^3 + 3x^2 + 4x + 4\end{align*}$ at $\begin{align*}x = 1\end{align*}$
      ::查找 x=1 时 m( x) = 3x3+3x2+4x+4 的图形的正切线的方程式
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  14. Find the equation of the tangent line to the graph of $\begin{align*}q(x) = -x^3 - 4x^2 + 4x + 3\end{align*}$ at $\begin{align*}x = -2\end{align*}$
      ::查找 q( x)\\\\\\\\\ x3\\ 4x2+4x+3 的图形的正切线的方程式 x\\ 2
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  15. Find the equation of the tangent line to the graph of $\begin{align*}t(x) = -4x^2 + 2x - 4\end{align*}$ at $\begin{align*}x = -1\end{align*}$
      ::在 x1 中查找 t(x) 4x2+2x-4 图形的正切线方程式方程式
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  16. Find the equation of the tangent line to the graph of $\begin{align*}h(x) = -4x^3 + 2x^2 - 3x + 3\end{align*}$ at $\begin{align*}x = -1\end{align*}$
      ::查找 h( x)\\\\\\ _4x3+2x2 -3x+3 x3 x3 xx+3 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
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  17. Find the equation of the tangent line to the graph of $\begin{align*}m(x) = x \end{align*}$ at $\begin{align*}x = 0\end{align*}$
      ::查找 x=0 时 m( x) =x 图形的正切线的方程式
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  18. Find the equation of the tangent line to the graph of $\begin{align*}s(x) = -3x^2 - 2x + 3 \end{align*}$ at $\begin{align*}x = 0\end{align*}$
      ::在 x=0 时查找 s(x)\\\\\\\%3x2-2-2x+3 图形的正切线方程式的方程式。
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  19. Find the equation of the tangent line to the graph of $\begin{align*}c(x) = -3\end{align*}$ at $\begin{align*}x = 0\end{align*}$
      ::在 x=0 中查找 c(x)\%3 图形的正切线的方程式。
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  20. Find the equation of the tangent line to the graph of $\begin{align*}b(x) = -5x^4 + 3x^3 - x^2 + 5x - 3 \end{align*}$ at $\begin{align*}x = -1\end{align*}$  
      ::查找 x\\\ 1 中的 b( x)\\\\ 5x4+3x3- x2+5x- 3 图形的正切线的方程式

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  ::回顾(答复)

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