Math, asked by hiteshnegiaaya3560, 6 days ago

find the maximum and minimum value of f(x)=4x^3+3x^2-18x+12

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Answered by alvisnand
0

Step-by-step explanation:

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Answered by dharanidx780
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Step-by-step explanation:

Enter a problem...

Calculus Examples

Popular Problems Calculus Find the Local

Maxima and Minima f(x)=4x^3-3x^2-18x+17

f(x)=3-3x2-18x+17

Find the first derivative of the function.

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12x2-6x-18

Find the second derivative of the function.

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f"(x)= 24x-6

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

0

and solve.

12x2-6x-18=0

Factor the left side of the equation.

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6(x+1)(2x-3)=0

Divide each term by

6

and simplify.

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2x2-x-3=0

Factor by grouping.

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(x+1)(2x-3)=0

If any individual factor on the left side of the equation is equal to

0

, the entire expression will be equal to

0

x+1=0. 2x-3=0

Set the first factor equal to

0

and solve.

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x=-1

Set the next factor equal to

0

and solve.

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x=32

The final solution is all the values that make

(x+1)(2x-3)=0

true.

x=-1,32

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.

24(-1)-6

Evaluate the second derivative.

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-30x=-1

is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

x=-1

is a local maximum

Find the y-value when

x=-1

.

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y= 28

Evaluate the second derivative at

x=32

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

24(32)-6

Evaluate the second derivative.

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30x=32

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

x= 32

is a local minimum

Find the y-value when

x=32

.

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y=-134

y

=

13

4

These are the local extrema for

f ( x)=4x3-3x2-18x+17. (-1,28)

is a local maxima

(32,-134)

is a local minima

image of graph

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