The graph of the derivative $f'(x)$ of a function $f(x)$ is shown in the figure below. If $g$ is a function defined for all $x$ by $g(x) = \left | 2f(x)+x^2 \right |$ and $f(-3)=-1$, how many local minima does the function $g$ have?


This is what I have done. From the figure, I can draw this table


We have $g(x)=\sqrt{( 2f(x)+x^2 )^2}$ so its derivative is $g'(x)=\frac{2(2f(x)+x^2)(2f'(x)+2x)}{2\sqrt{(2f(x)+x^2)^2}}$. I see that $2f'(-3)+2(-3)=0$ and $2f'(-1)+2(-1)=0$ so $g'(-3)=0$ and $g'(-1)=0$. I believe that there exists an $-3<x_0<-1$ such that $2f'(x_0)+2(x_0)=0$ which implies $g'(x_0)=0$. I'm stuck here, I don't know the sign of $g'(x)$ in any interval.

This is my prediction. We have $g(x)=\left | 2\left [ f\left ( x \right )-\left ( \frac{-x^2}{2} \right ) \right ] \right |$, consider the graph of the function $y=f(x)$ and the function $y=\frac{-x^2}{2}$. From the table above, I think that the graph of the function $y=f(x)$ is always "higher" (I don't know the exact word) than the graph of the function $y=\frac{-x^2}{2}$ for all $x$, which implies $\left [ f\left ( x \right )-\left ( \frac{-x^2}{2} \right ) \right ] >0$ for all $x$, so $g(x) = 2f(x)+x^2$ and its derivative is $g'(x)=2f'(x)+2x=2[f'(x)-(-x)]$. Now from the figure above, draw the graph of the function $y=-x$. We will see that the two graphs have 3 intersection points, which are $(-3,3)$, $(-1,1)$ and $(x_0,-x_0)$ where $-3<x_0<-1$, we notice that $f'(x)-(-x)>0$ for $-3<x<x_0$ (the graph of $y=f'(x)$ is higher) and $f'(x)-(-x)<0$ for $x_0<x<-1$ (the graph of $y=-x$ is higher). I draw this table


I conclude that $g$ has 2 local minima. If my prediction is right please show me how to prove this, or show me another way to finish this exercise. Thank you.


1 Answer 1


You're correct, but you can do this in a simpler way:

$$g'(x) = 2 \operatorname{sgn}(2 f(x) + x^2) (f'(x) + x)$$ where

$\operatorname{sgn}(x)$ is the sign of $x$.

We want $$g'(x) = 2 \operatorname{sgn}(2 f(x) + x^2) (f'(x) + x) = 0$$

Divide by $2 \operatorname{sgn}(2 f(x) + x^2)$ to get

$$f'(x) + x = 0$$ or equivalently $$x = -f'(x)$$

We can find all the points where $x = -f'(x)$ by looking at where $y = -x$ intersects the graph of $f'(x)$. We see that this occurs at $(-3,3)$, $(-1,1)$, and somewhere in between, and so we know that $g$ has 3 local extrema (not necessarily minima, as you noticed).

And then your logic about $f'(x)$ being "higher" (I don't know what the term would be either, but I understand what you mean) is correct, completing the solution by showing that minima exist at $x=-3$ and $x=-1$.

  • $\begingroup$ Can you show me the rigorous proof for the the fact that graph of the function $y=f(x)$ is always "higher" than the graph of the function $y=\frac{-x^2}{2}$ for all $x$? (May they have intersections?) $\endgroup$
    – Tinh Tran
    Nov 30, 2017 at 17:02
  • $\begingroup$ We can't prove this rigorously because the graph isn't rigorously defined. But eyeballing it gives the result. $\endgroup$
    – actinidia
    Nov 30, 2017 at 22:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .