Exam 2: Derivatives

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Find the derivative of the function. f(x)=14−4x+5x2f(x)=14-4 x+5 x^{2}

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10x−410 x-4

Complete the table by computing f(x)f(x) at the given values of x, accurate to five decimal places. Use the results to guess at the indicated limit, if it exists, to three decimal places. lim⁡x→5x+20−5x−5\lim _{x \rightarrow 5} \frac{\sqrt{x+20}-5}{x-5} x 4.9 4.99 4.999 5.001 5.01 5.1 f(x)

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x4.94.994.9995.0015.015.1f(x)0.10010.100010.10.10.099990.0999\begin{array}{|l|l|l|l|l|l|l|} \hline \text {x} &4.9 & 4.99 & 4.999 & 5.001 & 5.01 & 5.1 \\\hline \text {f(x)} &0.1001 & 0.10001 & 0.1 & 0.1 & 0.09999 & 0.0999\\\hline \end{array}

Use the definition of the derivative to find f′(3), where f(x)=x3−2x. f^{\prime}(3) \text {, where } f(x)=x^{3}-2 x \text {. }

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The position of a car is given by the values in the following table. tt t 0 1 2 3 4 (seconds) s 0 23.7 27 58.3 90.5 (meters) Find the average velocity for the time period beginning when t=2t=2 and lasting 2 seconds.

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Find the vertical asymptotes of the function. y=4x2+13x−4x2y=\frac{4 x^{2}+1}{3 x-4 x^{2}}

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If 1≤f(x)≤x2+6x+61 \leq f(x) \leq x^{2}+6 x+6 for all x find the limit. lim⁡x→−1f(x)\lim _{x \rightarrow-1} f(x)

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Use the graph of f(x)=x2−x−2x−2f(x)=\frac{x^{2}-x-2}{x-2} to guess at the limit lim⁡x→2x2−x−2x−2\lim _{x \rightarrow 2} \frac{x^{2}-x-2}{x-2} , if it exists.  Use the graph of  f(x)=\frac{x^{2}-x-2}{x-2}  to guess at the limit  \lim _{x \rightarrow 2} \frac{x^{2}-x-2}{x-2}  , if it exists.

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Find the vertical asymptotes of the function. y=4x2+13x−4x2y=\frac{4 x^{2}+1}{3 x-4 x^{2}}

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Find the limit lim⁡x→0+2+xx+49\lim _{x \rightarrow 0^{+}} \frac{2+\sqrt{x}}{\sqrt{x+49}} .

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Use the graph of the function to state the value of lim⁡x→0f(x)\lim _{x \rightarrow 0} f(x) if it exists. f(x)=11+52/xf(x)=\frac{1}{1+5^{2 / x}}

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If the tangent line to y=f(x) at (8,4)y=f(x) \text { at }(8,4) passes through the point (4,−32), find f′(8)(4,-32) \text {, find } f^{\prime}(8)

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Find the limit. lim⁡t→5t2−25t3−125\lim _{t \rightarrow 5} \frac{t^{2}-25}{t^{3}-125}

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If lim⁡x→3−f(x)=4.1\lim _{x \rightarrow 3^{-}} f(x)=4.1 , then if lim⁡x→3f(x)\lim _{x \rightarrow 3} f(x) exists, find the value where it exists.

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How would you define f(3)f(3) in order to make f continuous at 3 ? f(x)=x2−3x−7x−3f(x)=\frac{x^{2}-3 x-7}{x-3}

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Find an equation of the tangent line to the parabola y=5x3y=5 x^{3} at the point (−5,−145)(-5,-145) .

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Evaluate lim⁡h→0cot⁡(π4+h)−1h\lim _{h \rightarrow 0} \frac{\cot \left(\frac{\pi}{4}+h\right)-1}{h} .

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Estimate the value of the following limit by graphing the function f(x)=(5sin⁡x)(sin⁡πx)f(x)=\frac{(5 \sin x)}{(\sin \pi x)} . lim⁡x→05sin⁡xsin⁡πx\lim _{x \rightarrow 0} \frac{5 \sin x}{\sin \pi x} Round your answer correct to two decimal places.

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Evaluate the limit. lim⁡x→∞10x2−3x+17x2+3x−3\lim _{x \rightarrow \infty} \frac{10 x^{2}-3 x+1}{7 x^{2}+3 x-3}

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Determine where f is discontinuous. f(x)={−x if x<011−x if 0≤x<11(11−x)2 if x>11f(x)=\left\{\begin{array}{ccc}\sqrt{-x} & \text { if } & x<0 \\11-x & \text { if } & 0 \leq x<11 \\(11-x)^{2} & \text { if } & x>11\end{array}\right.

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Find the point at which the given function is discontinuous. f(x)={1x−8,x≠88,x=8f(x)=\left\{\begin{array}{ll}\frac{1}{x-8}, & x \neq 8 \\8, & x=8\end{array}\right.

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