Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.8 Intermediate Value Theorem - Exercises - Page 86: 13

Answer

The function $f(x)$ is continuous on $[0 ,1$ with $f(0)=1 \gt 0$ and $f(2)=-3 \lt 0$ Therefore, by the Intermediate Value Theorem, there is a $c \in (0, 2)$ such that $f(c)= 2^c+3^c -4^c =0 \implies 2^c+3^c =4^c$. So, the equation $2^x +3^x=-4^x $ has a solution $c$ in $(0,2)$.

Work Step by Step

We are given that $f(x)=2^x +3^x -4^x$ The function $f(x)$ is continuous on $[0 ,1$ with $f(0)=1 \gt 0$ and $f(2)=-3 \lt 0$ Therefore, by the Intermediate Value Theorem, there is a $c \in (0, 2)$ such that $f(c)= 2^c+3^c -4^c =0 \implies 2^c+3^c =4^c$. So, the equation $2^x +3^x=-4^x $ has a solution $c$ in $(0,2)$.
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