I have a quiz on "Maxima, minima, and the mean value theorem" and I don't even know which topic to apply to this question:
If $f''(x) \ge -1$, $x$ belongs to $(-15, 15)$, and $f'(1)=3$, find the interval over which $x$ is definitely increasing.
Please help :( this is making me so sad.
Note that $f$ is non decreasing iff $f'(x) \ge 0$.
Let $g(x) = f'(x)$ to keep the problem tick free inasmuch as possible.
We are given that $g'(x) \ge -1$, $g(1) = 3$.
By drawing a picture, we can surmise that for $x \ge 1$, the worst case is when $g'(x) = -1$ and so $g(x) \ge 0$ as long as $x \ge 4$. For $x < 1$, we can imagine a function for which $g(x) <0$, but $g$ grows quickly enough so that $g(1) = 3$. Hence our guess is that we can only guarantee $g(x) \ge 0$ for $x \in [1,4]$.
To validate the guess, note that for $x >1$ the mean value theorem shows that there is some $\xi \in [1,x]$ such that ${g(x)-3 \over x-1 } = g'(\xi) \ge -1$ from which we see that if $x \ge 4$ then $ g(x) \ge 0$. Furthermore, by considering $g(x) = 4 -x$ we see that $4$ is the best upper limit.
For $x<1$, consider the collection of functions $g_n(t) = 3-e^{-n(t-1)}$ and note that $\lim_{n \to \infty} g_n(x) = -\infty$ and $g'(t) \ge 0$ for all $t$. hence we see that $0$ is the best lower limit.