Proving $f(t)=f(0)$ using Mean Value Theorem

Question

Suppose that function $f: [0, \infty) \to \mathbb{R}$ is differentiable at every $t > 0$ and continuous at $= 0$. Show that if $f'(t) = 0$ for all $t > 0$ then $f'(t) = f'(0)$ for all $t ≥ 0$.

So, $f$ is a constant function. I will fix $t > 0$ and use the mean value theorem to $f$ on the interval $[0,t]$.

My attemp:
$f'(c)=\frac {f(t)-f(0)}{t-0}$
Based on what I read, $f'(c)$ should be zero. But, I am not sure how to prove it in this context.
$0= \frac {f(t)-f(0)}{t-0}$
$f(t) = f(0) $

I would appreciate your help and explanation. Thank you.

Answer 1

Since for all t > 0, f'(t) = 0,
exists k with for all t > 0, f(t) = k.
As f is continuous, f(0) = k. So f'(0) = 0.

Answer 2

As you mentioned $f$ is constant and also f is continuous at 0. So limit as $x→0$ is $f(0) $.

So the function is constant for all $t>=0$.

Now $f'(0)=0$ follows from definition of derivative