Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 3 - The Derivative - 3.3 Rates of Change - 3.3 Exercises - Page 158: 24

Answer

$f$ is increasing for $x=1$.

Work Step by Step

Recall that: The function$f$ is increasing if: $x \gt y \to f(x) \gt f(y)$ The function$f$ is decreasing if: $x \gt y \to f(x) \lt f(y)$ The instantaneous rate of change of $f$ at $x=1$ is: $$\lim\limits_{h \to 0}\frac{f(1+h)-f(1)}{h} \gt 0$$ It follows that the rate of change is also positive: $$\frac{f(1+h)-f(1)}{h} \gt 0$$ Since the step size $h$ is always positive it follows $h\gt 0$ or $1+h \gt 1$ so: $$f(1+h)-f(1) \gt 0$$ $$f(1+h)\gt f(1) $$ Since $1+h \gt 1$ we have $f(1+h) \gt f(1)$ it follows that $f$ is increasing for $x=1$.
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