When is a function strictly increasing?

Question

If I have a function, within an integral, how do I know if the function is strictly increasing for $x$ being greater than or equal to zero?

I thought maybe it might have something to do with the mean value theorem ?

Answer 1

A function $f : I \rightarrow X$ is strictly increasing if for all $x,y \in I$ such that $x < y$, we have that $f(x) < f(y)$. See here also for a condition on the derivative $f'$ of $f$ (assuming some conditions) which tells you if $f$ is increasing.

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With regards to your edits, and added image, the result which I linked to becomes more important. So try differentiating the function $F(x)$. As mentioned in the comments, you'll need to think about the Fundamental Theorem of Calculus. In general, this says that if we have (for $x \in [a,b]$) $$ F(x) = \int_a^x f(t) dt$$ then $F'(x)=f(x)$. So in our case, $F'(x) = e^x(x+1)^3$. Indeed, for $x \geq 0$, $F'(x) > 0$. Note that in the above link, we only get that this is increasing. For strictly increasing, we invoke this condition. But for our case, $F'(x) \neq 0$ for all $x \geq 0$, so we have a strictly increasing function.

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We can also do this more directly using the MVT; we can pick $a,b \in [0, x]$ such that $ a < b$. Then $F$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Then by the MVT, there exists a $c \in (a,b)$ such that $$ F'(c) = \frac{F(b) - F(a)}{b-a}.$$ As we've already shown, $F'(c) > 0$, and we know that $b-a>0$. Hence $F(b) - F(a) > 0 \implies F(b) > F(a) $. We can do this for all $a,b \in [0, x]$ with $a < b$, so $F$ is strictly increasing.

Answer 2

Hint: Differentiate the function. What do you get? What is the sign of the resulting function?