* SOLUTION 1 :* The domain of *f* is all *x*-values. Now determine a sign chart for the first derivative, *f*' :

*f*'(*x*) = 3*x*^{2} - 6*x*

= 3*x* (*x* - 2)

= 0

for *x*=0 and *x*=2 . See the adjoining sign chart for the first derivative, *f*' .

Now determine a sign chart for the second derivative, *f*'' :

*f*''(*x*) = 6*x* - 6

= 6 (*x* - 1)

= 0

for *x*=1 . See the adjoining sign chart for the second derivative, *f*'' .

Now summarize the information from each sign chart.

FROM *f*' :

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

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

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

*f* has a relative minimum at *x*=2 , *y*=-4 .

FROM *f*'' :

*f* is ()
for *x*>1 ;

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

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

OTHER INFORMATION ABOUT *f* :

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

Click HERE to return to the list of problems.

* SOLUTION 2 :* The domain of *f* is all *x*-values. Now determine a sign chart for the first derivative, *f*' :

*f*'(*x*) = 4*x*^{3} - 12*x*^{2}

= 4*x*^{2} (*x* - 3)

= 0

for *x*=0 and *x*=3 . See the adjoining sign chart for the first derivative, *f*' .

Now determine a sign chart for the second derivative, *f*'' :

*f*''(*x*) = 12*x*^{2} - 24*x*

=12*x* (*x* - 2)

= 0

for *x*=0 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 *x*>3 ;

*f* is (
)
for *x*<0 and 0<*x*<3 ;

*f* has an absolute minimum at *x*=3 , *y*=-27 .

FROM *f*'' :

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

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

*f* has inflection points at *x*=0 , *y*=0 and *x*=2 , *y*=-16 .

OTHER INFORMATION ABOUT *f* :

If *x*=0 , then *y*=0 so that *y*=0 is the *y*-intercept. If *y*=0 , then
*x*^{4}-4*x*^{3}=*x*^{3}(*x*-4)=0 so that *x*=0 and *x*=4 are the *x*-intercepts. There are no vertical or horizontal asymptotes since *f* is a polynomial. See the adjoining detailed graph of *f* .

Click HERE to return to the list of problems.

* SOLUTION 3 :* The domain of *f* is all *x*-values. Now determine a sign chart for the first derivative, *f*' . Using the ordinary product rule, we get

*f*'(*x*) = *x*^{3} 2 (*x*-2) + 3*x*^{2} (*x*-2)^{2}

(Factor out *x*^{2} and (*x*-2) .)

= *x*^{2} (*x*-2) [ 2*x* + 3(*x*-2) ]

= *x*^{2} (*x*-2) [ 5*x*-6 ]

= 0

for *x*=0 , *x*= 6/5 , and *x*=2 . See the adjoining sign chart for the first derivative, *f*' .

Now determine a sign chart for the second derivative, *f*'' . Using the triple product rule, we get

*f*''(*x*) = 2*x*(*x*-2)[5*x*-6] + *x*^{2}(1)[5*x*-6] + *x*^{2}(*x*-2)[5]

(Factor out *x* .)

= *x* [ 2(*x*-2)(5*x*-6) + *x* (5*x*-6) + 5*x*(*x*-2) ]

= *x* [ 2(5*x*^{2}-16*x*+12) + 5*x*^{2}-6*x* + 5*x*^{2}-10*x* ]

= *x* [ 20*x*^{2}-48*x*+24 ]

= 4*x* [ 5*x*^{2}-12*x*+6 ]

= 0

for *x*=0 , and (using the quadratic formula)
. See the adjoining sign chart for the second derivative, *f*'' .

Now summarize the information from each sign chart.

FROM *f*' :

*f* is ()
for *x*<0 , 0<*x*<6/5 , and *x*>2 ;

*f* is (
)
for 6/5 <*x*<2 ;

*f* has a relative maximum at *x*=6/5 ,
;

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

FROM *f*'' :

*f* is ()
for
and
;

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

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

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*-2)^{2}=0 so that *x*=0 and *x*=2 are the *x*-intercepts. There are no vertical or horizontal asymptotes since *f* is a polynomial. See the adjoining detailed graph of *f* .

Click HERE to return to the list of problems.

* SOLUTION 4 :* The domain of *f* is all *x*-values. Now determine a sign chart for the first derivative, *f*' . Using the quotient rule, we get

= 0

for *x*= 1 , and *x*=-1 . 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* and (*x*^{2}+1) .)

= 0

for *x*=0 ,
, and
. See the adjoining sign chart for the second derivative, *f*'' .

Now summarize the information from each sign chart.

FROM *f*' :

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

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

*f* has an absolute maximum at *x*=1 , *y*=2 ;

*f* has an absolute minimum at *x*=-1 , *y*=-2 .

FROM *f*'' :

*f* is ()
for
and
;

*f* is ()
for
and
;

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

OTHER INFORMATION ABOUT *f* :

If *x*=0 , then *y*=0 so that *y*=0 is the *y*-intercept. If *y*=0 , then
so that 4*x*=0 and *x*=0 is the *x*-intercept. There is a horizontal asymptote since

= 0 .

Thus, the line *y* = 0 is a a horizontal asymptote for the graph of *f* . See the adjoining detailed graph of *f* .

Click HERE to return to the list of problems.

* SOLUTION 5 :* The domain of *f* is all *x*-values EXCEPT *x*=2 , because of division by zero. Now determine a sign chart for the first derivative, *f*' . Using the quotient rule, we get

= 0

for *x*= 1 , and *x*=3 . In addition, note that *f*' is NOT DEFINED at *x*=2 . 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 4 and (*x*-2) .)

(Divide out a factor of (*x*-2) .)

= 0

for NO *x*-values. However, note that *f*'' is NOT DEFINED at *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 *x*<1 and *x*>3 ;

*f* is (
)
for 1<*x*<2 and 2<*x*<3 ;

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

*f* has a relative minimum at *x*=3 , *y*=9 .

FROM *f*'' :

*f* is ()
for *x*>2 ;

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

*f* has no inflection points.

OTHER INFORMATION ABOUT *f* :

If *x*=0 , then *y*=0 so that *y*=0 is the *y*-intercept. If *y*=0 , then
so that
2*x*^{2}-3*x*=*x*(2*x*-3)=0 . Thus, *x*=0 and *x*=3/2 are *x*-intercepts. There is no horizontal asymptote since

,

and

.

Remember, a horizontal asymptote exists only if the limit to
or
is a finite number. Now check for a vertical asymptote by computing one-sided limits at the zero of the denominator, i.e., at *x*=2 . Thus,

(The numerator approaches 2 and the denominator is a positive number approaching 0 .)

,

and

(The numerator approaches 2 and the denominator is a negative number approaching 0 .)

.

This shows that the line *x* = 2 is a vertical asymptote for the graph of *f* . Remember, if EITHER of these one-sided limits is
or ,
a vertical asymptote exists. See the adjoining detailed graph of *f* .

Click HERE to return to the list of problems.

* SOLUTION 6 :* The domain of *f* is all *x*-values EXCEPT *x*=2 and *x*=-2 , because of division by zero. Now determine a sign chart for the first derivative, *f*' . Using the quotient rule, we get

(Factor out 2 and (*x*-4) .)

= 0

for *x*= 1 , and *x*=4 . In addition, note that *f*' is NOT DEFINED at *x*=2 and *x*=-2 . 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 and (*x*^{2}-4) .)

(Divide out a factor of (*x*^{2}-4) .)

= 0 ,

so that
-2*x*^{3} + 15*x*^{2} -24*x* + 20 = 0 . To solve this equation use Newton's method or an equation solver like that found on a TI85 graphing calculator, getting a single real solution
. 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 *x*<-2 , -2<*x*<1 , and *x*>4 ;

*f* is (
)
for 1<*x*<2 and 2<*x*<4 ;

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

*f* has a relative minimum at *x*=4 , *y*=0 .

FROM *f*'' :

*f* is ()
for *x*<-2 and 2<*x*<5.70 ;

*f* is ()
for -2<*x*<2 and *x*>5.70 ;

*f* has an inflection point at
,
.

OTHER INFORMATION ABOUT *f* :

If *x*=0 , then *y*=-4 so that *y*=-4 is the *y*-intercept. If *y*=0 , then
so that (*x*-4)^{2}=0 . Thus, *x*=4 is the *x*-intercept. There is a horizontal asymptote since

= 1 .

Thus, the line *y*=1 is a horizontal asymptote for the graph of *f* . Now check for vertical asymptotes by computing one-sided limits at the zeroes of the denominator, i.e., at *x*=2 and at *x*=-2 . Thus,

(The numerator approaches 4 and the denominator is a positive number approaching 0 .)

,

and

(The numerator approaches 4 and the denominator is a negative number approaching 0 .)

.

This shows that the line *x* = 2 is a vertical asymptote for the graph of *f* . Remember, if EITHER of these one-sided limits is
or ,
a vertical asymptote exists. Now check at *x*=-2 . Thus,

(The numerator approaches 36 and the denominator is a negative number approaching 0 .)

,

and

(The numerator approaches 36 and the denominator is a positive number approaching 0 .)

.

This shows that the line *x* = -2 is a vertical asymptote for the graph of *f* . See the adjoining detailed graph of *f* .

Click HERE to return to the list of problems.