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Q23E

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Essential Calculus: Early Transcendentals

Found in: Page 216

Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

If \(f(1) = 10\) and \({f^\prime }(x) \ge 2\) for \(1 \le x \le 4\), how small can \(f(4)\) possibly be?

The smallest value of \(f(4)\) is \(16\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

The values are \(f(1) = 10\) and \({f^\prime }(x) \ge 2\) for \(1 \le x \le 4\).

Step 2: Concept of Mean value theorem

“Let \(f\) be a function that satisfies the following hypothesis:

1. \(f\) is continuous on the closed interval \(\left( {a\;,{\rm{ }}b} \right)\).

2. \(f\) is differentiable on the open interval \(\left( {a,b} \right)\).

Then, there is a number \(c\) in \(\left( {a\;,{\rm{ }}b} \right)\) such that \({f^\prime }(c) = \frac{{f(b) - f(a)}}{{b - a}}\).

Or, equivalently, \(f(b) - f(a) = {f^\prime }(c)(b - a)\).”

Step 3: Calculation to find the smallest possible value of \(f(4)\)

From the given conditions, it is observed that the values of \(a = 1\) and \(b = 4,f(a)\) is represented by \(f(1) = 10\), \({f^\prime }(c)\) is given by \({f^\prime }(x) \ge 2\) and to find possible \(f(4)\) which is represented by \(f(b)\).

Substitute the given values in \({f^\prime }(c) = \frac{{f(b) - f(a)}}{{b - a}}\).

\(\begin{aligned}{c}{f^\prime }(c) &= \frac{{f(4) - f(1)}}{{4 - 1}}\\{f^\prime }(c) &= \frac{{f(4) - 10}}{3}\\f(4) - 10 &= {f^\prime }(c)(3)\\f(4) &= 3{f^\prime }(c) + 10\end{aligned}\)

Simplify further to check the smallest possible value of \(f(4)\) and substitute \({f^\prime }(x) \ge 2\) for \({f^\prime }(c)\).

\(\begin{aligned}{c}f(4) \ge 3(2) + 10\\f(4) \ge 6 &+ 10\\f(4) \ge 16\end{aligned}\)

Thus, the smallest value of \(f(4)\) is \(16\).

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