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Show that if $f : \mathbb{R} \longrightarrow \mathbb{R}$ is differentiable and $f'(x) \geq x$, then $f(x) \leq f(0) + \frac{1}{2}x^2$ for all $x \leq 0$.

I'm wondering whether this statement can be proved using Taylor's Theorem. Since we're given that $f$ is only one times differentiable, we can use, $f(x) = P_{0}(x_0) + R_1(x),$ $x_0 = 0$ and $x \leq 0$. This gives:

\begin{align} f(x) = f(0) + f'(c)x, \quad\quad c \in [x, 0], \end{align}

Using the assumption, we can write:

\begin{align} f(x) & = f(0) + f'(c)x \\ & \geq f(0) + cx \\ & \geq f(0) + x^2 \\ \end{align}

But using this method, I not only miss out on the factor of $2$ but also get the reverse inequality. Any suggestions?

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  • $\begingroup$ it is for x negative, so you get $-\int_x^0$ and the reverse inequality. $\endgroup$
    – zwim
    May 6, 2017 at 17:20
  • $\begingroup$ Oh, right. I see that now. The remainder term, using the form that I have used above, will be negative and not positive. But I'm still not sure where to get the factor of 2 from. $\endgroup$ May 6, 2017 at 17:22
  • $\begingroup$ a primitive of $t$ is $t^2/2$ $\endgroup$
    – zwim
    May 6, 2017 at 17:23
  • $\begingroup$ I'm explicitly trying not to use the integral form of the remainder term. I'm trying to use the form given by the expression: $R_n(x) = \frac{f^{n+1}(c)}{(n+1)!}(x - x_0)^{n+1}$ $\endgroup$ May 6, 2017 at 17:24
  • $\begingroup$ For $n = 0$, this gives me $R_0(x) = f'(c)x$. Since $n+1$ is odd, the remainder term will be negative, but I'm still not sure how to get the factor of 2 using this approach. $\endgroup$ May 6, 2017 at 17:26

1 Answer 1

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Try using

$$f(x) = f(0) + \int_0^x f'(t) \text{ d}t$$

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  • $\begingroup$ I'm not what, if any, difference would this make. It'd be great if you could elaborate on your answer. You're using the integral form of the remainder term. I'm not sure what difference will this make assuming one makes the same manipulations as I did above. $\endgroup$ May 6, 2017 at 17:16
  • $\begingroup$ Try substituting $f'(t) = t$ into the integral above $\endgroup$
    – user156213
    May 6, 2017 at 17:23

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