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