What is the extreme f(x)=-2x+3ln(4x),(1,5) ?

The extreme f(x)=-2x+3ln(4x),(1,5) is Maximum(3/2 ,-3+3ln(6))

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extreme f(x)=-2x+3ln(4x),(1,5)

extreme points

f (x) = - 2 x + 3 ln (4 x), (1, 5)

Maximum (\frac{3}{2}, - 3 + 3 ln (6))

Solution steps

Find the critical points:

x = \frac{3}{2}

Domain of

- 2 x + 3 ln (4 x), (1, 5) : 1 < x < 5

Find intervals:

Increasing

: 1 < x < \frac{3}{2},

Decreasing

: \frac{3}{2} < x < 5

Plug the extreme point

x = \frac{3}{2}

into

- 2 x + 3 ln (4 x) \Rightarrow y = - 3 + 3 ln (6)

Maximum (\frac{3}{2}, - 3 + 3 ln (6))

f (x) = 3 x^{2} - 30 x + 72

f (x) = x^{3} + 3 x^{2} - 9

- x^{3} + 2 x^{2}

f (x) = x ln (x^{2})

f (x) = \frac{x ^{2} - 11 x + 67}{x - 9}

What is the extreme f(x)=-2x+3ln(4x),(1,5) ?
The extreme f(x)=-2x+3ln(4x),(1,5) is Maximum(3/2 ,-3+3ln(6))