Question

Find the minimum and maximum value of x⁴-8x²

Answer 1

Answer:

Correct option is

B

25,−39

Differentiate the function f(x)=3x

4

−8x

3

+12x

2

−48x+25 with respect to x,

f

′

(x)=12x

3

−24x

2

+24x−48

Put f

′

(x)=0,

12x

3

−24x

2

+24x−48=0

(x−2)(12x

2

+24)=0

(x−2)12(x

2

+2)=0

x−2=0

x=2

Or,

x

2

=−2

x=

2

i

But

2

i is an imaginary number, so the value to be considered is x=2.

Now, substitute the value of x and the end points in the given function.

f(0)=3(0)

4

−8(0)

3

+12(0)

2

−48(0)+25

=25

f(2)=3(2)

4

−8(2)

3

+12(2)

2

−48(2)+25

=−39

f(3)=3(3)

4

−8(3)

3

+12(3)

2

−48(3)+25

=10

Therefore, the maximum value is 25 and the minimum value is −39.

Step-by-step explanation:

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