* SOLUTION 7 :* The domain of *f* is all *x*-values. Now determine a sign chart for the first derivative, *f*' . Then

*f*'(*x*) = 1 - 3 (1/3) *x*^{1/3-1}

= 1 - *x*^{-2/3}

= 0

when
*x*^{2/3} - 1 = 0 . Thus,
*x*^{2/3} = 1 so that
and
*x* = 1 or *x* = -1 . In addition, note that *f*' is NOT DEFINED at *x*=0 . To avoid using a calculator, it is convenient to use numbers like
*x*= -8, -1/8, 1/8, and 8 as "test points" to construct the adjoining sign chart for the first derivative, *f*' .

Now determine a sign chart for the second derivative, *f*'' . Beginning with

*f*'(*x*)= 1 - *x*^{-2/3} ,

we get

*f*''(*x*)= 0 - (-2/3) *x*^{-2/3 - 1}

= (2/3) *x*^{-5/3}

= 0

for NO *x*-values . However, note that *f*'' is NOT DEFINED at *x*=0 . To avoid using a calculator, it is convenient to use numbers like *x*= -1 and *x*=1 as "test points" to construct the adjoining sign chart for the second derivative, *f*'' .

Now summarize the information from each sign chart.

FROM *f*' :

*f* is ()
for *x*<-1 and *x*>1 ;

*f* is (
)
for -1<*x*<0 and 0<*x*<1 ;

*f* has a relative maximum at *x*=-1 , *y*=2 ;

*f* has a relative minimum at *x*=1 , *y*=-2 .

FROM *f*'' :

*f* is ()
for *x*>0 ;

*f* is ()
for *x*<0 ;

*f* has an inflection point at *x*=0 , *y*=0 .

OTHER INFORMATION ABOUT *f* :

If *x*=0 , then *y*=0 so that *y*=0 is the *y*-intercept. If *y*=0 , then
*x* - 3*x*^{1/3} = *x*^{1/3} ( *x*^{2/3} - 3 ) = 0 so that *x*^{1/3}=0 or
*x*^{2/3} = 3 . Thus,
or
, so that
and *x*=0 are the *x*-intercepts. There are no horizontal or vertical asymptotes. See the adjoining detailed graph of *f* .

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* SOLUTION 8 :*
The domain of *f* is all *x*-values. Now determine a sign chart for the first derivative, *f*' . Beginning with

= (5/2)*x*^{2/3} - *x*^{2/3} *x*^{1}

= (5/2)*x*^{2/3} - *x*^{2/3+1}

= (5/2)*x*^{2/3} - *x*^{5/3}

we get

*f*'(*x*) = (5/2)(2/3)*x*^{2/3-1} - (5/3)*x*^{5/3-1}

= (5/3) *x*^{-1/3} - (5/3)*x*^{2/3}

= 0

when *x*=1 . In addition, note that *f*' is NOT DEFINED at *x*=0 . See the adjoining sign chart for the first derivative, *f*' .

Now determine a sign chart for the second derivative, *f*'' . Beginning with

*f*'(*x*)= (5/3)*x*^{-1/3} - (5/3) *x*^{2/3} ,

we get

*f*''(*x*)= (5/3)(-1/3) *x*^{-1/3-1} - (5/3)(2/3) *x*^{2/3-1}

= (-5/9) *x*^{-4/3} - (10/9) *x*^{-1/3}

= 0

when 1+2*x*=0 or *x*= -1/2 . In addition, note that *f*'' is NOT DEFINED at *x*=0 . See the adjoining sign chart for the second derivative, *f*'' .

Now summarize the information from each sign chart.

FROM *f*' :

*f* is ()
for 0<*x*<1 ;

*f* is (
)
for *x*<0 and *x*>1 ;

*f* has a relative maximum at *x*=1 , *y*= 3/2 ;

*f* has a relative minimum at *x*=0 , *y*=0 .

FROM *f*'' :

*f* is ()
for *x*< -1/2 ;

*f* is ()
for -1/2<*x*<0 and *x*>0 ;

*f* has an inflection point at *x*=-1/2 ,
.

OTHER INFORMATION ABOUT *f* :

If *x*=0 , then *y*=0 so that *y*=0 is the *y*-intercept. If *y*=0 , then
*x*^{2/3} ( 5/2 - *x* ) = 0 so that *x*^{2/3}=0 or
5/2 - *x* = 0 . Thus, *x*=0 and *x* = 5/2 are the *x*-intercepts. There are no horizontal or vertical asymptotes. See the adjoining detailed graph of *f* .

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* SOLUTION 9 :*
The domain of *f* is artificially restricted to be
. Now determine a sign chart for the first derivative, *f*' . Begin with

= 0 .

Then

so that

or

.

A well-known angle *x* in the second quadrant with

and

is
. A well-known angle *x* in the fourth quadrant with

and

is
. It is convenient to use
and
as test points in the first derivative
when constructing the adjoining sign chart for the first derivative, *f*' .

Now determine a sign chart for the second derivative, *f*'' . Begin with

= 0 .

Then

so that

or

.

A well-known angle *x* in the first quadrant with

and

is
. A well-known angle *x* in the third quadrant with

and

is
. It is convenient to use
and
as test points in the second derivative
when constructing the adjoining sign chart for the second derivative, *f*'' .

Now summarize the information from each sign chart.

FROM *f*' :

*f* is ()
for
and
;

*f* is (
)
for
;

*f* has a relative maximum at
,
;

*f* has an absolute maximum at
, *y*= 2 ;

*f* has a relative minimum at *x*=0 ,

*f* has an absolute minimum at
, *y*=-2 .

FROM *f*'' :

*f* is ()
for
and
;

*f* is ()
for
;

*f* has inflection points at
, *y*=0 and
, *y*=0 .

OTHER INFORMATION ABOUT *f* :

If *x*=0 , then
, so that
is the *y*-intercept. If *y*=0 , then
and
(This is the same equation used to solve *f*''(*x*)=0 .) so that
and
are the *y*-intercepts. There are no horizontal or vertical asymptotes. See the adjoining detailed graph of *f* .

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* SOLUTION 10 :* Since
the domain of *f* is
in order to insure that we not take the square root of a negative number . Now determine a sign chart for the first derivative, *f*' . Using the product rule, we get

= 0

for *x*= 0 ,
,and
. In addition, note that *f*' is NOT DEFINED at *x*=2 and *x*=-2 since these numbers lead to division by zero. See the adjoining sign chart for the first derivative, *f*' .

Now determine a sign chart for the second derivative, *f*'' . Beginning with

and using the quotient rule, we get

(Factor out 2*x* .)

= 0 .

Then
so that *x*= 0 ,
, or
. HOWEVER,
and
are not in the domain of function *f* so only *x*=0 solves *f*'(*x*)=0 . In addition, note that *f*'' is NOT DEFINED at *x*=2 and *x*=-2 . See the adjoining sign chart for the second derivative, *f*'' .

Now summarize the information from each sign chart.

FROM *f*' :

*f* is ()
for
;

*f* is (
)
for
and
;

*f* has an absolute maximum at
,
;

*f* has a relative maximum at *x*=-2 , *y*=0 ;

*f* has an absolute minimum at
,
;

*f* has a relative minimum at *x*=2 , *y*=0 .

FROM *f*'' :

*f* is ()
for -2<*x*<0 ;

*f* is ()
for 0<*x*<2 ;

*f* has an inflection point at *x*=0 , *y*=0 .

OTHER INFORMATION ABOUT *f* :

If *x*=0 , then *y*=0 , so that *y*=0 is the *y*-intercept. If *y*=0 , then
so that *x*=0 , *x*=2 , and
*x*=-2 are the *y*-intercepts. There are no horizontal or vertical asymptotes. See the adjoining detailed graph of *f* .

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* SOLUTION 11 :*
First note that
*y* = *A x*^{3} + 6*x*^{2} - *Bx* is a polynomial so that it is infinitely-differentiable. The first derivative is

*y*'= 3*Ax*^{2} + 12*x* - *B* ,

and the second derivative is

*y*''= 6*Ax* + 12 .

If the graph of *y* has a maximum value at *x*= -1, then *y*'(1)=0 so that
3*A*(-1)^{2} + 12(-1) - *B* = 0 or

(Equation 1)

3*A* - *B* = 12 .

If the graph of *y* has an inflection point at *x*=1, then *y*''(1)=0 so that
6*A*(1) + 12 = 0 or

*A* = -2 .

Substituting this value for *A* into Equation 1, we get
3(-2) - *B* = 12 or

*B* = -18 .

Thus, assuming that *y*'(-1)=0 and *y*''(1)=0 , the polynomial must be of the form

*y* = -2*x*^{3} + 6*x*^{2} + 18*x*

with first derivative

*y*' = -6*x*^{2} + 12*x* + 18

and second derivative

*y*'' = -12*x* + 12 .

Check to see if *x*=-1 determines a maximum value for *y* . Using the Second Derivative Test for Extrema, we have that *y*'(-1)=0 and
*y*''(-1) = -12(-1) + 12 = 24 > 0 ! Thus, *x*=-1 determines a MINIMUM value for *y* , NOT a maximum value . This problem has NO SOLUTION.

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