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Found in: Page 216

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# 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 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 $$16$$.