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It is given that a function $f$ is differentiable everywhere, and $f(0)=5$. If $f(x)<5$ for all nonzero $x$ then what is the value of $f'(0)$?.

Now I see that $0$ is a point of maximum of the function, which is differentiable at $0$, which means that $f'(0)$ must be zero. Is this the correct answer? I am conflicted because the book says it's not. There is a chance of a misprint though.

Not Euler
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2 Answers2

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You're correct.

Consider $\frac{f(h)-f(0)}{h}$; this is precisely $\frac{f(h)-5}{h}$, which is always negative when $h > 0$, and positive when $h < 0$. Therefore taking the limit as $h \to 0$ from above, the limit must be $\leq 0$; and as $h \to 0$ from below, the limit must be $\geq 0$; so (since the limit does indeed exist because $f$ is differentiable) the limit must be $0$.

Patrick Stevens
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In the situation you are describing the point $x_0=0$ is the maximum of the function. Hence using Fermat's theorem (See Link) you have that $f'(0)=0$.

Yanko
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