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Suppose that $f: \mathbb{R} \rightarrow \mathbb{R}$ is a continuous function on the interval $[-3, 3]$ and a differentiable function in the interval $(-3,3)$ such that for every $x$ in the interval, $f’(x) \leq 2$. If $f(-3)=7$, then $f(3)$ is at most __________
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Given that $f’(X) \leq 2$ and $f(-3) = 7$,

As maximum slope is positive and we need value of $f(x)$ at $3$ which is on right side of $-3$,  we can assume $f(x)$ as a straight line with slope $2$. It will give us the correct result.

Let $f(x) = 2x + b,$

$f(-3) = -6 + b = 7 \Rightarrow b = 13$

$f(x) = 2x + 13$

$f(3)_{\max} = 6 + 13 = 19.$


Correct method would be using Mean Value Theorem. Above method will work only if you can analyze the cases correctly and can assume the $f(x)$ without any loss of accuracy, otherwise you are very prone to commit a mistake that way.

Using Mean Value Theorem:

$f’(c) = \frac{f(b) – f(a)}{b -a }$

     $\Rightarrow f’(c) = \frac{f(3) – f(-3)}{3 -(-3)} =  \frac{f(3) – 7}{6} $

     $\Rightarrow f(3) = 6f’(c) +7 $

As $f’(c)\leq 2$,

     $\Rightarrow f(3) \leq 6*2 +7 = 19.$

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5 votes
5 votes

The answer is $19$. Check the image below.

 

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0 votes
So we are given the max angle as 2 .

We know tan(theta) = height/base

 

So, 2= height / (3 - (-3))

Therefore : height = 2*6 = 12

And we know the level is at .. 7 so answe is 12+7 = 19
Answer:

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