1.) Find the x-coordinate of the absolute maximum for the function f(x)=(4+6ln(x

Question

1.) Find the x-coordinate of the absolute maximum for the function

f(x)=(4+6ln(x))/(x) ,x>0.

x-coordinate of absolute maximum =

2.) Find the x-coordinate of the absolute minimum for the function

f(x)=5xln(x)?9x, x>0.

x-coordinate of absolute minimum =

3.)

Find the absolute maximum and absolute minimum values of the function

f(x)=x^3-12x^2-27x+7

over each of the indicated intervals.

(a) Interval = [?2,0].

(b) Interval = [1,10].

(c) Interval = [?2,10].

Explanation / Answer

1.) f'(x) = ((6/x)*x - (4+6ln(x))/x^2 = 0

So 6 - 4 - 6ln(x) = 0

6ln(x) = 2

ln(x) = 1/3

So x = e^(1/3)

2.) f'(x) = (5+5ln(x) - 9 = 0

5ln(x) = 4

ln(x) = 4/5

So x = e^(4/5)

3.) f'(x) = 3x^2 - 24x - 27 = 0

x^2 - 8x - 9 = 0

(x-9)(x+1) = 0

So x = -1 , 9

f''(x) = 6x - 24

f''(-1) = 6(-1) - 24 = -30 < 0

So At x = -1 , absolute maximum

(A1.) Absolute maximum = f(-1) = (-1)^3 - 12(1) - 27(-1) +7 = 21

(A2.) Absolute minimum = f(-2) = (-2)^3 - 12(-2)^2 - 27(-2) + 7 = 5

f''(9) = 6(9) - 24 = 30 > 0

So At x = 9 , absolute minimum

(B1.) Absolute maximum = f(1) = (1)^3 - 12(1) - 27(1) +7 = -31

(B2.) Absolute minimum = f(9) = (9)^3 - 12(9)^2 - 27(9) + 7 = -479

(C1.) Absolute maximum = f(-1) = (-1)^3 - 12(1) - 27(-1) +7 = 21

(C2.) Absolute minimum = f(9) = (9)^3 - 12(9)^2 - 27(9) + 7 = -479

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