$\vert f'(a) \vert < k$,prove that there exists $I$, $a\in I$ that $\vert f'(x)\vert<k$, $f \in C^1$

Question

so here's the question:

$\vert f'(a) \vert < k$,prove that there exists $I$, $a\in I$ that $\vert f'(x)\vert<k$, $f \in C^1$, since the first derivative is continuous, I tried using the continuity limit definition, however I wasn't able to finish it, and I tried using the MVT, Didn't help either.

Intuitively to me it sounds kinda obvious, Since we're working on $\mathbb{R}$ and since It is continuous in the neighbourhood of $a$, we're bound to have some interval $I$ no matter how small it is, that satisfies that condition.

Any help is appreciated! Thanks.

Answer 1

Since $f$ is $C^1$, $f'$ is continuous.

Let $\epsilon = k - |f'(a)|$.

There exists $\delta > 0$ with the property that $|x-a| < \delta$ implies $|f'(x) - f'(a)| < \epsilon$.

Let $I = (a-\delta,a+\delta)$.

Then $$x \in I \implies |x-a| < \delta \implies |f'(x) - f'(a)| < \epsilon \implies |f'(x)| < |f'(x) - f'(a)| + |f'(a)|$$ so that $$x \in I \implies |f'(x)| < (k - |f'(a)|) + |f'(a)| = k.$$