Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.1 - Derivatives of Polynomials and Exponential Functions - 3.1 Exercises - Page 181: 57


1) Find the derivative of $y$, or $y'$ 2) Consider the equation $y'(x)=2$. Prove that $[y'(x)-2]\gt0$ for all $x$. That means $[y'(x)-2]\ne0$, or $y'(x)\ne2$. Therefore, there is no tangent line with slope 2.

Work Step by Step

$$y=f(x)=2e^x+3x+5x^3$$ 1) First, find the derivative of $y$ $$y'=2\frac{d}{dx}(e^x)+3\frac{d}{dx}(x)+5\frac{d}{dx}(x^3)$$ $$y'=2e^x+15x^2+3$$ 2) A tangent line to the curve $y$ at point $A(x,y)$ that has slope 2 must have $y'(x)=2$. In other words, $$2e^x+15x^2+3=2$$ $$2e^x+15x^2+1=0$$ We see that $e^x\gt0$ for all $x$ and $15x^2\ge0$ for all $x$. That means $2e^x+15x^2+1\gt0$ for all $x$. Therefore, $2e^x+15x^2+1\ne0$. In other words, $y'(x)\ne2$. Which means there is no tangent line that has slope 2.
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.