Find the relative extrema, if any, of the function. Use the Second Derivative Te

Question

Find the relative extrema, if any, of the function. Use the Second Derivative Tes i applicable (If an answer does not exist, enter DNE f(x) 2x4 8x 4 relative maximum relative minimum Find the relative extrema, if any, of the function. Use the Second Derivative Test, if applicable (If an answer does not exist, enter DNE.) relative maximum relative minimum Find the relative extrema, if any, of the function. Use the Second Derivative Test, if applicable. (If an answer does not exist, enter DNE.) f(x 5 sine x 0

Explanation / Answer

(1)

given function is

f(x)=2x4-8x+4

f'(x)=8x3-8

f''(x)=24x2

for finding maximum or minimum value we equate 1st derivative to 0,then

8x3-8=0

x=1

putting this value in f''(x)

f''(x)=24>0

therefore minimum value exist for this function and that is at point x=1

minimum value is

f(1)=-2

no maximum value(because function is increasing function)

relative maximum=DNE

relative minimum=-2

(2)

given function is

f(t)=2t+1/t

f'(t)=2-1/t2

f''(t)=-2/t3

again for finding maximum or minimum we have to equate 1st derivative with 0,so

2-1/t2=0

t=1/sqrt(2)

putting this in f''(t)

f''(1/sqrt(2))=-4sqrt(2)<0

so there exist a maximium value for this function at point t=1/sqrt(2)

maximium value for this function is

f(1/sqrt(2))=2sqrt(2)

and there is no minimum value for this function as function is decreasing function

relative maximium=2sqrt(2)

relative minimum=DNE

(3)

given function is

f(x)=5sin2x 0<x<3pi/2

f'(x)=10sinx cosx

f'(x)=5sin(2x)

f''(x)=10cos(2x)

again on comparing

5sin(2x)=0

2x=pi 0<x<3pi/2

x=pi/2

so putting this value in f''(x)

f''(pi/2)=-10<0

so there is maximum value for this function is at x=pi/2 and value is

f(pi/2)=5

and for minimum value we can say that

sin2x>=0

means its minimum value is 0

therefore function minimum value is 0

relative maximium=5

relative minimum=0

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