### SOLUTIONS TO GRAPHINGOF FUNCTIONS USING THE FIRST AND SECOND DERIVATIVES

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

f'(x) = 3x2 - 6x

= 3x (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) = 6x - 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 .

If x=0 , then y=0 so that y=0 is the y-intercept. If y=0 , then x3-3x2=x2(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 .

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

f'(x) = 4x3 - 12x2

= 4x2 (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) = 12x2 - 24x

=12x (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 .

If x=0 , then y=0 so that y=0 is the y-intercept. If y=0 , then x4-4x3=x3(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 .

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) = x3 2 (x-2) + 3x2 (x-2)2

(Factor out x2 and (x-2) .)

= x2 (x-2) [ 2x + 3(x-2) ]

= x2 (x-2) [ 5x-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) = 2x(x-2)[5x-6] + x2(1)[5x-6] + x2(x-2)[5]

(Factor out x .)

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

= x [ 2(5x2-16x+12) + 5x2-6x + 5x2-10x ]

= x [ 20x2-48x+24 ]

= 4x [ 5x2-12x+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 , .

If x=0 , then y=0 so that y=0 is the y-intercept. If y=0 , then x3(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 .

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 2x and (x2+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 , .

If x=0 , then y=0 so that y=0 is the y-intercept. If y=0 , then so that 4x=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 .

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.

If x=0 , then y=0 so that y=0 is the y-intercept. If y=0 , then so that 2x2-3x=x(2x-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 .

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 (x2-4) .)

(Divide out a factor of (x2-4) .)

= 0 ,

so that -2x3 + 15x2 -24x + 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 , .

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 .