Give the value of $x$ where the function $$f(x) = x^3 + 3x^2 − 24x$$ has a local maximum.
I tried graphing the problem, but I'm still not sure exactly how to find the local max. Could you please help?
What is the local maximum value of the function $f(x) = x^3 + 3x^2 − 24x$?
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calculus
3 Answers
1
Do note that $f'(x)=0$ doesn't instantly mean a maximum or a minimum, the derivative must change signs, or, while $f'(x)=0$, $f''(x)<0$ or $f''(x)>0$.
An example of this is $f(x)=x^3$.
At $x=0$ there is a horizontal tangent, but not a local maximum or minimum, because the first derivative does not change signs.
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It is known the local maximum occurs when $f'(x)=0, f''(x)<0$.
Note that $f'(x)=3x^2+6x-24=3(x+4)(x-2)$ and $f''(x)=6x+6$. Notice $$f''(2)>0, f''(-4)<0$$
Thus the local maximum occurs when $x=-4$.
2
To find the critical points of the equation, you have to set the first derivative $f'(x)=0$. Once you have the critical numbers, you may check to see whether they represent a minimum, maximum or an inflection point. From here, take the second derivative to show the concavity, which will reveal whether you have a maximum or a minimum. For a maximum, you would be looking for the x value where $f"(x)<0$.
In this case, start by setting $f'(x)=0=3x^2+6x-24$. Then, factor out the 3, which leaves you equivalently with $x^2+2x-8=0$. By factoring, you get $(x+4)(x-2)=0$, leaving you with critical numbers {-4, 2}.
From here, take these two points and simply plug them into the second derivative, and check which leaves you with a negative number.
By plugging in our critical points, we see that $$f"(2)>0$$ and $$f"(-4)<0$$
Hence, the maximum occurs at $x=-4$