## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

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

#### Answer

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.

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