Math, asked by shanmugam5699, 1 year ago

Find the minimal value of f(x)=3x^4-8x^3+6x^2-12 on [-3,3]

Answers

Answered by pinquancaro
5

Consider the given polynomial

f(x) = 3x^4-8x^3+6x^2-12

Let us determine the first derivative, we get

f'(x) = 12x^3-24x^2+12x

Let f'(x) = 0

12x^3-24x^2+12x = 0

12x(x^2-2x+1)=0

12x (x^2-x-x+1))=0

12x (x(x-1)-(x-1)) = 0

x = 0 and x = 1

Now, let us find second derivative, we get

f''(x) =36x^2-48x+12

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 deepchatterjeepune
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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