Question

1) The function f(x)=2x3−33x2+108x+3f(x)=2x3-33x2+108x+3 has one local minimum and one local maximum. Use a graph of the function to estimate these local extrema.

This function has a local minimum at x =     with output value =

and a local maximum at x =     with output value =

2) The function f(x)=2x3−24x2+42x+7 has one local minimum and one local maximum. Use a graph of the function to estimate these local extrema.

This function has a local minimum at x =     with output value =

and a local maximum at x =     with output value=

Answer 1

1)

f(x) =2x^3−33x^2+108x +3

Find derivative as

f'(x) = 6x² -66x +108

To find critical pint, solve f'(x) =0

6x² -66x +108 =0

6(x² -11x +18) =0

6(x-2)(x-9) =0

x =2, 9

Find second derivative as

f"(x) = 12x -66

Here

f"(2) < 0, hence x = 2 is absolute maximum

f"(9) >0, hence x = 2 is absolute minimum

Now

f(2) =103

f(9)= -240

hence

This function has a local minimum at x = 9 with output value = -240

and a local maximum at x = 2 with output value = 103

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2)

This function has a local minimum at x = 7 with output value = -189

and a local maximum at x =1 with output value = 27