Let the function f:[-7,0]R be continuous on [-7,0] and differentiable on (-7,0).

Question

Let the function $f:[-7,0]	o\mathbb{R}$ be continuous on $[-7,0]$ and differentiable on $(-7,0)$. If $f(-7)=-3$ and $f'(x)\le2$ for all $x\in(-7,0)$, then for all such functions $f$, $f'(-1)+f(0)$ lies in the interval:

Accepted Answer

Using Lagrange's Mean Value Theorem in [--7, --1]

$${{f\left( { - 1} \right) - f\left( { - 7} \right)} \over { - 1 - \left( { - 7} \right)}}$$ = f'(c1)

As f'(x) $$ \le $$ 2 then f'(c1) $$ \le $$ 2

$$ \therefore $$ $${{f\left( { - 1} \right) - f\left( { - 7} \right)} \over { - 1 - \left( { - 7} \right)}}$$ $$ \le $$ 2

$$ \Rightarrow $$ $${{f\left( { - 1} \right) + 3} \over 6}$$ $$ \le $$ 2

$$ \Rightarrow $$ f(-1) $$ \le $$ 9

Using Lagrange's Mean Value Theorem in [--7, 0]

$$ \Rightarrow $$ $${{f\left( 0 \right) - f\left( { - 7} \right)} \over {0 - \left( { - 7} \right)}}$$ $$ \le $$ 2

$$ \Rightarrow $$ f(0) $$ \le $$ 11

$$ \therefore $$ f(-1) + f(0) $$ \le $$ 11 + 9

$$ \Rightarrow $$ f(-1) + f(0) $$ \le $$ 20

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