the maximum and minimum values of f(x)= 4x^3 + 3x^2 -6x + 5 are?
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Answers
f(x) = 4x^3 + 3x^2 -6x + 5
f'(x) = 12x^2 + 6x - 6
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 = 6/12
x = 1/2
The final solution is all the values that make 12x^2 + 6x - 6 = 0 true
x = -1, 1/2
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= (1/2).
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(1/2) + 6
= 12 + 6
= 6
x=(1/2) 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=1/2
f(x) = 4x^3 + 3x^2 -6x + 5
F(1/2) = 4/8 + 3/4 - 6/2 + 5
F(1/2) = 1/2 + 3/4 - 3 + 5
F(1/2) = (2+3+8)/4 = 13/4
Y = 13/4
so,
These are the local extrema
for f(x) = 4x^3 + 3x^2 -6x + 5
(-1 , 10) is a local maxima
(1/2 , 13/4) is a local minima