Maths-

General

Easy

Question

# The maximum and minimum values of are

- 8,7/2
- 10/13
- 3,5/7
- 2,8/7

Hint:

### Hint: Here we have to find the maximum and minimum values off(x) = 4x3 + 3x2 -6x + 5. For that, find the first and second derivate and then local minimum and local maximum value. And put these value in the function to find the actual minimum and maximum value.

## The correct answer is: 10/13

### Here we have to find the maximum and minimum value.

Firstly , we have,

f(x) = 4x^{3} + 3x^{2} -6x + 5

first derivative,

f'(x) = 12x^{2} + 6x – 6

second derivative,

F"(x) = 24x + 6

To find the local maximum and minimum values of the function, set the derivative equal to 0 and solve.

12x^{2} + 6x - 6 = 0

12x^{2} + 12x - 6x - 6 = 0

12x(x + 1) - 6(x + 1) = 0

(x + 1) (12x - 6) = 0

so,

(x + 1) = 0

x = -1

&

(12x - 6) = 0

x =

x =

The final solution is all the values that make 12x^{2} + 6x - 6 = 0 true

x = -1,

Evaluate the second derivative at x= (-1).

If the second derivative is positive, then this is a local minimum.

If it is negative, then this is a local maximum.

24x + 6

=24(-1) + 6

=-24 + 6

= -18

x=(-1) is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

Find the y-value when x=(-1)

f(x) = 4x^{3} + 3x^{2} -6x + 5

F(-1) = 4(-1) + 3(+1) - 6(-1) + 5

F(-1) = -4 + 3 + 6 + 5 = 10

Y = 10

Evaluate the second derivative at x= ().

If the second derivative is positive, then this is a local minimum.

If it is negative, then this is a local maximum.

24x + 6

= 24() + 6

= 12 + 6

= 6

x=() is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

Find the y-value when x=

f(x) = 4x^{3} + 3x^{2} -6x + 5

F() = + - + 5

F() = + - 3 + 5

F() = =

Y =

Therefore, the maximum and minimum number is 10 ,

In this question, you have to find the maximum and minimum value of the function. Firstly find the local minimum and local maximum value put for that , put the value of 1st derivative in 2nd derivative to find local maximum and minimum value. If the 2nd derivative is positive then it is minimum else maximum. Put the same value to the main equation and find the minimum and maximum value.

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