University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.1 - Tangents and the Derivative at a Point - Exercises - Page 119: 49

Answer

The graph is plotted below. All the details are explained in the Work step by step.
1537630707

Work Step by Step

$$f(x)=x^3+2x\hspace{1cm}x_0=0$$ a) Here, since $x_0=0$, we will plot $y=f(x)$ over the interval $-1/2\le x\le3$. The graph of $y=f(x)$ is represented by the red curve. b) and c) With $x_0=0$: $$q(h)=\frac{f(h)-f(0)}{h}=\frac{f(h)-(0^3+2\times0)}{h}=\frac{f(h)}{h}$$ The graph of $q(h)$ is also plotted and represented by the purple line. As $h\to0$, from the graph, we see that $q$ approaches $2$. Therefore, $$\lim_{h\to0}q(h)=2$$ d) For $x_0=0$: $$y=f(0)+q(h)\times(x-0)=0+q(h)\times x=q(h)\times x$$ - With $h=3$: $$y=q(3)\times x=\frac{f(3)}{3}\times x=\frac{(3^3+2\times3)x}{3}=\frac{33x}{3}=11x$$ This secant line is represented by the black line. - With $h=2$: $$y=q(2)\times x=\frac{f(2)}{2}\times x=\frac{(2^3+2\times2)x}{2}=\frac{12x}{2}=6x$$ This secant line is represented by the blue line. - With $h=1$: $$y=q(1)\times x=\frac{f(1)}{1}\times x=\frac{(1^3+2\times1)x}{1}=\frac{3x}{1}=3x$$ This secant line is represented by the green line. - The tangent line at $x_0=0$: The tangent line at $x=0$ will have this form: $$y=\lim_{h\to0}q(h)x+m=2x+m$$ To find m, we use the fact that the tangent line is at $x=0$: $$2\times0+m=f(0)=0$$ $$m=0$$ So the tangent line at $x_0$ is $y=2x$. It is represented by the orange line.
Small 1537630707
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.