How do you find the relative extrema of a function with only its first derivative and not the function itself?

One of the problems on my practice asks to find the relative extrema of $$f(x)$$ if $$f'(x)=x(x+5)^3(x-2)^2$$ where $$f'(x)$$ is the first derivative of $$f(x)$$. I know that the critical points of $$f(x)$$ are $$-5$$, $$0$$, and $$2$$ and that a sign chart showing $$f'(x)$$ can be used to determine whether there are relative extrema at each of these points. However, I'm unsure of how to find the actual values of these relative extrema because there is no initial function, namely $$f(x)$$, to plug these x-values back into. Any help would be appreciated!

Edit: Also, I should have clarified this. The reason I know that the y-values are being asked for is because when the x-values are being asked for the practice use the wording "find the values of the domain where relative extrema occur." For this question, it asks for the actual relative extrema of f(x) for a set of questions, but I'm not sure how to find the y-values for this specific scenario.

• It is impossible to determine the precise values of the function from the derivative alone. To see why this is the case, suppose you know the function $f(x)$. If you take the function $g(x)=f(x)+10$, then the derivative of $g(x)$ is identical to the derivative of $f(x)$, and it will have extremes in the same $x$-values... but will take different $y$-values at those points. How do you know, given $f'(x)$, that you "had" $f(x)$ and not $g(x)$? Or $f(x)-17$? Or $f(x)+\pi$? You cannot. It is impossible to determine the values of $f(x)$ solely from knowledge of what $f'(x)$ is. Commented Aug 24, 2023 at 0:57
• This makes sense. Thank you so much! I spent over an hour trying to figure this out but was getting nowhere. Now I know why!
– SSS
Commented Aug 24, 2023 at 1:08

It is not possible, given a function $$f'(x)$$ which is the derivative of an unknown function and nothing else, to determine the values of the function $$f(x)$$ at any point, let alone at the extremes.

(If you have more information, such as the value of $$f(x)$$ at a single point, then it is possible; that would be what is usually called an "initial value problem" in Differential Equations. But you need that additional information.)

To see why, let's look at an example: suppose I am told

The derivative of $$f(x)$$ is $$4x^3-3x^2$$. Find the value of $$f(x)$$ at $$x=0$$.

Well, you could say "Ah, I can find a function with $$f'(x)=4x^3-3x^2$$; I'm just lucky enough to notice that $$f(x)=x^4 - x^3$$ works. Now I plug in $$x=0$$, and find $$f(0)=0$$."

Well, sure. The problem is that there are infinitely many other functions that also have derivative equal to $$4x^3-3x^2$$. For example, $$f_1(x)=x^4-x^3+1$$. And $$f_2(x)=x^4-x^3+2$$. And $$f_{\pi}(x)=x^4-x^3+\pi$$. And for any real number $$r$$, the function $$f_r(x) = x^4-x^3+r$$ has derivative equal to $$4x^3-3x^2$$. Each of them will have a different value at $$x=0$$: $$f_r(0)=r$$. Which one was the original? It is impossible to say.

Intuitively, it should be clear why all these functions work: because the graph of $$g(x) = f(x)+c$$, with $$c$$ a constant, is just a vertical shift of the graph of $$f(x)$$. That means that the tangents at any given value of $$x$$ to the graphs of $$g$$ and of $$f$$ will have the same slope... so you get the same derivative.

What is perhaps not so obvious is that those functions are the only functions whose derivative is $$4x^3-3x^2$$. Every function $$h(x)$$ with $$h'(x)=4x^3-3x^2$$ is equal to $$h(x) = x^4-x^3+c$$ for some constant $$c$$. This is a consequence of the Mean Value Theorem.

This last fact is why if you happen to also know the value of $$f(x)$$ at some point, in addition to knowing $$f'(x)$$, then you can use the two pieces of information to find the value of $$f(x)$$ at any point. If in addition to knowing that $$f'(x) = 4x^3-3x^2$$, I also know the value of $$f(2)$$, then I can plug $$2$$ into $$x^4-x^3+c$$ and solve for $$c$$. For instance, if I know that $$f(2)=0$$, then $$0 = f(2) = 2^4 - 2^3 + c = 16-8+c = 8+c$$, so that $$f(x)$$ must be the function $$f(x)=x^4-x^3-8$$, and now I can find the value of $$f$$ at any point.

Finding the x-values which you appear to know how to do well is sometimes sufficient for "Finding the extrema".

Depending on the exact wording of the question the y-values may not be required.

• Sorry, I should have clarified. The reason I know that the y-values are being asked for is because when the x-values are being asked for the practice use the wording "find the values of the domain where relative extrema occur." For this question, it asks for the actual relative extrema of $f(x)$, but I'm not sure how to find the y-values.
– SSS
Commented Aug 24, 2023 at 0:42
• You haven't learned integration yet? One could expand the polynomial. Integration or finding the antiderivative is fairly simple. However, it adds, +C, a constant of integration. Assuming C = 0 is generally not recommended. Commented Aug 24, 2023 at 0:47
• No, we have not been taught integration yet so that is not a way I can approach the problem for now. Is there a way to find the y-values without integration?
– SSS
Commented Aug 24, 2023 at 0:48