Find the minimal value of f(x)=3x^4-8x^3+6x^2-12 on [-3,3]
Answers
Answered by
5
Consider the given polynomial
Let us determine the first derivative, we get
Let f'(x) = 0
)=0
= 0
x = 0 and x = 1
Now, let us find second derivative, we get
f''(x) =
Now, f''(0) = 0-0+12 = 12
f''(1) = 36-48+12 = -12+12 = 0
If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.
Since, second derivative is positive at x = 0
So, x = 0 is the minimal value of the given function.
Answered by
3
Answer:
-12
Step-by-step explanation:
I agree with Pinquancaroupto second last point i.e f(x) attains minimum value at x=0 but the actual minimum value is
f(0) = -12
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