Suppose that it is given to you that f\'(x) = (x + 3) (11 - x) (17 - x) The the

Question

Suppose that it is given to you that f'(x) = (x + 3) (11 - x) (17 - x) The the first extremum (from the left) for f(x) occurs at x = The function f(x) has a relative at this point. The second relative extremum (from the left) for f (x) occurs at x = The function f(x) has a relative at this point. The third relative extremum (from the left) for f(x) occurs at x = The function f(x) has a relative at this point. The first inflection point (from the left) for f(x) occurs at x = The second inflection point (from the left) for f(x) occurs at x =

Explanation / Answer

give f '(x)=(x+3)(4-x)(x-15)

f '(x)=-(x+3)(x-4)(x-15)

singn chart:

x...............-3..................4...................15................

f '(x)....+...............-...................+..................-................

the first relative extremum (from the left ) for f(x) occurs at x =-3

the function f(x) has a relative maximum at this point

the second relative extremum (from the left ) for f(x) occurs at x =4

the function f(x) has a relative minimum at this point

the third relative extremum (from the left ) for f(x) occurs at x =15

the function f(x) has a relative maximum at this point

f '(x)=(x+3)(4-x)(x-15)

f '(x)=-x3+16x2-3x-180

f "(x)=-3x2+32x-3

for inflection point f "(x)=0

-3x2+32x-3 =0

by quadratic formula

x=(16-247)/3,x=(16-247)/3

the first inflection point (from the left ) for f(x) occurs at x =(16-247)/3=0.094589

the second inflection point (from the left ) for f(x) occurs at x =(16+247)/3=10.572

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