Say I have a function $f(x) = e^x -2$ with $x \in [0, 1]$. How do I show that $f(0) < f(x) < f(1)$ $\forall x \in [0,1]$.

Question

I know this is quite a basic question, but at this current time I can't think of a proof.

Here's the question.

Could someone please point me in the right direction to writing a proof for this?

Do you know the notion of monotonic function? The notion of derivative? What have you try? — mathcounterexamples.net, Apr 26 '20 at 14:43

score 0 · Accepted Answer · answered Apr 26 '20 at 14:58

You need to show that the function is increasing on $[0,1]$. To show that obtain the derivative of $f(x)$ and show that $f'(x)>0 \;\; \forall x \in [0,1]$.

Clearly, $f'(x)=e^x$ and we have that $f(x)>0\;\;\forall x \in \mathbb{R}$. Hence, $f'(x)>0\;\;\forall x \in [0,1]$. Hence, the function is (strictly) increasing function. So, we have $f(0)<f(x)<f(1)$.

Say I have a function $f(x) = e^x -2$ with $x \in [0, 1]$. How do I show that $f(0) < f(x) < f(1)$ $\forall x \in [0,1]$.

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