what is the local maximum value of function f(x)=x³+3x²_24x
Answers
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Answer:
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=3x2+6x−24. Then, factor out the 3, which leaves you equivalently with x2+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
Answer:
80
Step-by-step explanation:
f(x)= x^3+3x^2-24x
f'(x)= 3x^2+6x-24
for critical points put f'(x)=0
3x^2+6x-24=0
x^2+2x-8=0
x^2+4x-2x-8=0
x(x+4)-2(x+4)
x= 2 and -4
plug and get minima and Maxima
SHORTCUT
find f"(x)= 6x+6
for Maxima f"(x)<0
thus for x=-4 the function has the maximum value
-64+3(16)-24(-4)
-64+48+96
80
hope it helps u
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