Find, if any, the absolute maxima and minima of the function given on the interv

Question

Find, if any, the absolute maxima and minima of the function given on the interval [15, infinity).f(x) = 3 / x-2 Determine the intervals upon which the given function is increasing or decreasing. f(x) = x4 - 16x3 Determine the intervals upon which the given function is increasing or decreasing. f(x) = 2x + 4 Find, if any, the relative maxima and relative minima of the function. f(x) = x + 4/x +2 Find the interval on which f(x) = x2 +2x- 2 is increasing and the interval upon which it is decreasing. Find the relative extrema for f(x) = x4 + 2x2

Explanation / Answer

f(x) = 3/(x-2)

For maxima or minima f'(x) = 0

f'(x) = -3/(x-2)^2 which can not be zero

so we have to check end points

f(5) = 3/(5-2) = 1

Absolute maxima : 1 at x = 5

Absolute minima: none

f(x) = x4 - 16x3

=> For increasing or decreasing f'(x) > 0 and f'(x)<0 respectively

=> f'(x) = 4x3 - 48x2

for increasing f'(x) = 4x3 - 48x2 > 0

=> (x-1) > 0 ====> x in (1,inf)

for decreasing f'(x) = 4x3 - 48x2 < 0

=> (x-1) < 0=======> x in (-inf,1)

f(x) = sqrt(2x + 4)

=> f'(x) = 1/2 * (2x + 4)^(3/2)

=>for increasing f'(x) > 0 =>  1/2 * (2x + 4)^(3/2) > 0

=> x > -2 ===========> x in (-2,inf)

=>for decreasing f'(x) < 0 =>  1/2 * (2x + 4)^(3/2) < 0

=> x<-2=============> x in (-inf,-2)

f(x) = x + 4/x + 2

=>f'(x) = 1 - 4/x2

=> f'(x) = 0 = 1 - 4/x2

=> x = 2, x = -2

f''(x) = 8 / x3

=> f''(2) = 1 > 0 so minima at x = 2 ==========> f(2) = 2 + 4/2 + 2 = 6

f''(-2) = -1 < 0 so maxima at x = -2 ==========> f(-2) = -2 - 2 +2 = -2

f(x) = x2 + 2x - 2

=> f'(x) = 2x + 2

For increasing f'(x) > 0 => 2x+2>0

=============> x in (-1,inf)

For dec f'(x) < 0

=============> x in (-inf,-1)

f(x) = x2 + 2x4

=> f'(x) = 2x + 8x3

f'(x) = 0 =>

2x + 8x3 = 0

=> (4x2 + 1) x = 0

=> x = 0

f''(x) = 2 + 24x2

=> f''(0) = 2 > 0

=========> Relative minima at x = 0 and f(0) = 0

=========> Relative maxima: none

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