Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 1 - Section 1.5 - Inverse Functions and Logarithms - 1.5 Exercises - Page 66: 59

Answer

See solution

Work Step by Step

For every value of x, there is a different value of y, so this is a one to one function. As a more formal proof, the derivative of the function is $f'(x)=\frac{3x^2+2x+1}{2\cdot\sqrt{x^3+x^2+x+1}}$, where $2\cdot\sqrt{x^3+x^2+x+1}\neq0$. $3x^2+2x+1$ has $D=b^2-4ac=-8$, which is smaller than 0. This means the slope is never equal to 0 and is strictly increasing or decreasing. From the graph, we can see the function is strictly increasing.
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