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

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) x1/3-1

= 1 - x-2/3

= 0

when x2/3 - 1 = 0 . Thus, x2/3 = 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 .

If x=0 , then y=0 so that y=0 is the y-intercept. If y=0 , then x - 3x1/3 = x1/3 ( x2/3 - 3 ) = 0 so that x1/3=0 or x2/3 = 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 .

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

= (5/2)x2/3 - x2/3 x1

= (5/2)x2/3 - x2/3+1

= (5/2)x2/3 - x5/3

we get

f'(x) = (5/2)(2/3)x2/3-1 - (5/3)x5/3-1

= (5/3) x-1/3 - (5/3)x2/3

= 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) x2/3 ,

we get

f''(x)= (5/3)(-1/3) x-1/3-1 - (5/3)(2/3) x2/3-1

= (-5/9) x-4/3 - (10/9) x-1/3

= 0

when 1+2x=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 , .

If x=0 , then y=0 so that y=0 is the y-intercept. If y=0 , then x2/3 ( 5/2 - x ) = 0 so that x2/3=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 .

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 .

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 .

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 2x .)

= 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 .

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 .

SOLUTION 11 : First note that y = A x3 + 6x2 - Bx is a polynomial so that it is infinitely-differentiable. The first derivative is

y'= 3Ax2 + 12x - B ,

and the second derivative is

y''= 6Ax + 12 .

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

(Equation 1)

3A - B = 12 .

If the graph of y has an inflection point at x=1, then y''(1)=0 so that 6A(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 = -2x3 + 6x2 + 18x

with first derivative

y' = -6x2 + 12x + 18

and second derivative

y'' = -12x + 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.