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 ($\uparrow$) for x<0 and x>2 ;

f is ( $\downarrow$) 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 ($\cup$) for x>1 ;

f is ($\cap$) 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 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 .

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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 ($\uparrow$) for x>3 ;

f is ( $\downarrow$) for x<0 and 0<x<3 ;

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

FROM f'' :

f is ($\cup$) for x<0 and x>2 ;

f is ($\cap$) 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 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 .

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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) $x= \displaystyle{ -(-12) \pm \sqrt{ (-12)^2 - 4(5)(6) } \over 2(5) } = \displ... ...style{ 12 \pm 2\sqrt{6 } \over 10 } = \displaystyle{ 6 \pm \sqrt{6 } \over 5 }$ . See the adjoining sign chart for the second derivative, f'' .

Now summarize the information from each sign chart.

FROM f' :

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

f is ( $\downarrow$) for 6/5 <x<2 ;

f has a relative maximum at x=6/5 , $y=(6/5)^3(-4/5)^2 \approx 1.11$ ;

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

FROM f'' :

f is ($\cup$) for $0<x<\displaystyle{6-\sqrt{6} \over 5 }$ and $x>\displaystyle{6+\sqrt{6} \over 5 }$ ;

f is ($\cap$) for x<0 and $${6-\sqrt{6} \over 5 }<x<\displaystyle{6+\sqrt{6} \over 5 }$$ ;

f has inflection points at x=0 , y=0 and $x=\displaystyle{6-\sqrt{6} \over 5 } \approx 0.71$ , $y \approx 0.60$ and $x=\displaystyle{6+\sqrt{6} \over 5 } \approx 1.69$ , $y \approx 0.46$ .

OTHER INFORMATION ABOUT f :

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 .

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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

$f'(x) = \displaystyle{ (x^2+1)(4) - (4x)(2x) \over (x^2+1)^2 }$

$= \displaystyle{ 4x^2+4 - 8x^2 \over (x^2+1)^2 }$

$= \displaystyle{ 4 - 4x^2 \over (x^2+1)^2 }$

$= \displaystyle{ 4 (1-x^2) \over (x^2+1)^2 }$

$= \displaystyle{ 4 (1-x)(1+x) \over (x^2+1)^2 }$

= 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

$f'(x) = \displaystyle{ 4 - 4x^2 \over (x^2+1)^2 }$

and using the quotient rule, we get

$f''(x) = \displaystyle{ (x^2+1)^2 (-8x)-(4 - 4x^2)2(x^2+1)(2x) \over (x^2+1)^4 }$

(Factor out 2x and (x²+1) .)

$= \displaystyle{ 2x(x^2+1) [ -4(x^2+1)- 2(4-4x^2) ] \over (x^2+1)^4 }$

$= \displaystyle{ 2x [ -4x^2-4- 8+8x^2 ] \over (x^2+1)^3 }$

$= \displaystyle{ 2x [ 4x^2-12 ] \over (x^2+1)^3 }$

$= \displaystyle{ 2x (4) [ x^2-3 ] \over (x^2+1)^3 }$

$= \displaystyle{ 8x (x-\sqrt{3})(x+\sqrt{3}) \over (x^2+1)^3 }$

= 0

for x=0 , $x= \sqrt{3}$ , and $x= -\sqrt{3}$ . See the adjoining sign chart for the second derivative, f'' .

Now summarize the information from each sign chart.

FROM f' :

f is ($\uparrow$) for -1<x<1 ;

f is ( $\downarrow$) 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 ($\cup$) for $-\sqrt{3}<x<0$ and $x> \sqrt{3}$ ;

f is ($\cap$) for $x<-\sqrt{3}$ and $0<x<\sqrt{3}$ ;

f has inflection points at x=0 , y=0 and $x= -\sqrt{3}$ , $y=-\sqrt{3}$ and $x= \sqrt{3}$ , $y= \sqrt{3}$ .

OTHER INFORMATION ABOUT f :

If x=0 , then y=0 so that y=0 is the y-intercept. If y=0 , then $${ 4x \over x^2 + 1 }=0$$ so that 4x=0 and x=0 is the x-intercept. There is a horizontal asymptote since

$${ \lim_{ x \to \pm \infty } \ { 4x \over x^2 + 1 } } = \displays... ...er x^2 + 1 } { \displaystyle{1 \over x^2} \over \displaystyle{1 \over x^2} } }$$

$= \displaystyle{ \lim_{ x \to \pm \infty } \ { \displaystyle{ 4 \over x } \over 1 + \displaystyle{1 \over x^2 } } }$

$= \displaystyle{ 0 \over 1 + 0 }$

= 0 .

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

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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

$f'(x) = \displaystyle{ (x-2)(4x-3) - (2x^2-3x)(1) \over (x-2)^2 }$

$= \displaystyle{ 4x^2-8x-3x+6-2x^2+3x \over (x-2)^2 }$

$= \displaystyle{ 2x^2-8x+6 \over (x-2)^2 }$

$= \displaystyle{ 2(x^2-4x+3) \over (x-2)^2 }$

$= \displaystyle{ 2(x-1)(x-3) \over (x-2)^2 }$

= 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

$f'(x) = \displaystyle{ 2x^2-8x+6 \over (x-2)^2 }$

and using the quotient rule, we get

$f''(x) = \displaystyle{ (x-2)^2(4x-8) - (2x^2-8x+6) 2(x-2) \over (x-2)^4 }$

$= \displaystyle{ (x-2)^2 4(x-2) - 2(x^2-4x+3) 2(x-2) \over (x-2)^4 }$

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

$= \displaystyle{ 4(x-2) [(x-2)(x-2) - (x^2-4x+3) ] \over (x-2)^4 }$

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

$= \displaystyle{ 4 [x^2-4x+4 - x^2+4x-3) ] \over (x-2)^3 }$

$= \displaystyle{ 4 [1 ] \over (x-2)^3 }$

$= \displaystyle{ 4 \over (x-2)^3 }$

= 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 ($\uparrow$) for x<1 and x>3 ;

f is ( $\downarrow$) 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 ($\cup$) for x>2 ;

f is ($\cap$) 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 $${ 2x^2-3x \over x-2 }=0$$ so that 2x²-3x=x(2x-3)=0 . Thus, x=0 and x=3/2 are x-intercepts. There is no horizontal asymptote since

$${ \lim_{ x \to \infty } \ { 2x^2-3x \over x-2 } } = \displaystyl... ...2-3x \over x-2 } { \displaystyle{1 \over x} \over \displaystyle{1 \over x} } }$$

$= \displaystyle{ \lim_{ x \to \infty } \ { 2x - 3 \over 1 - \displaystyle{2 \over x } } }$

$= \displaystyle{ \infty \over 1 - 0 }$

$= \infty$ ,

and

$${ \lim_{ x \to -\infty } \ { 2x^2-3x \over x-2 } } = \displaysty... ...2-3x \over x-2 } { \displaystyle{1 \over x} \over \displaystyle{1 \over x} } }$$

$= \displaystyle{ \lim_{ x \to -\infty } \ { 2x - 3 \over 1 - \displaystyle{2 \over x } } }$

$= \displaystyle{ -\infty \over 1 - 0 }$

$= -\infty$ .

Remember, a horizontal asymptote exists only if the limit to $\infty$ or $-\infty$ 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,

$${ \lim_{ x \to 2^{+} } \ { 2x^2-3x \over x-2 } } = { 2(2)^2-3(2) \over 0^{+} }$$

$= \displaystyle{ 2 \over 0^{+} }$

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

$= + \infty$ ,

and

$${ \lim_{ x \to 2^{-} } \ { 2x^2-3x \over x-2 } } = { 2(2)^2-3(2) \over 0^{-} }$$

$= \displaystyle{ 2 \over 0^{-} }$

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

$= -\infty$ .

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 $\infty$ or $-\infty$, a vertical asymptote exists. See the adjoining detailed graph of f .

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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

$f'(x) = \displaystyle{ (x^2-4)2(x-4) - (x-4)^2(2x) \over (x^2-4)^2 }$

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

$= \displaystyle{ 2(x-4) [(x^2-4) - x(x-4) ] \over (x^2-4)^2 }$

$= \displaystyle{ 2(x-4) [ x^2-4 - x^2 + 4x ] \over (x^2-4)^2 }$

$= \displaystyle{ 2(x-4) [ 4x-4 ] \over (x^2-4)^2 }$

$= \displaystyle{ 2(x-4)(4) [ x-1 ] \over (x^2-4)^2 }$

$= \displaystyle{ 8(x-4) ( x-1 ) \over (x^2-4)^2 }$

= 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

$f'(x)= \displaystyle{ 8( x^2 - 5x + 4 ) \over (x^2-4)^2 }$

and using the quotient rule, we get

$f''(x) = \displaystyle{ (x^2-4)^2 8(2x-5) - 8( x^2 - 5x + 4 ) 2(x^2-4)(2x) \over (x^2-4)^4 }$

(Factor out 2 and (x²-4) .)

$= \displaystyle{ 2(x^2-4) [ 4(x^2-4)(2x-5) - 16x( x^2 - 5x + 4 ) ] \over (x^2-4)^4 }$

(Divide out a factor of (x²-4) .)

$= \displaystyle{ 2 [ 4(2x^3-5x^2-8x+20) - 16x^3 +80x^2 -64x ] \over (x^2-4)^3 }$

$= \displaystyle{ 2 [ 8x^3-20x^2-32x+80 - 16x^3 +80x^2 -64x ] \over (x^2-4)^3 }$

$= \displaystyle{ 2 [ -8x^3 + 60x^2 -96x + 80 ] \over (x^2-4)^3 }$

$= \displaystyle{ 2 (4) [ -2x^3 + 15x^2 -24x + 20 ] \over (x^2-4)^3 }$

$= \displaystyle{ 8 [ -2x^3 + 15x^2 -24x + 20 ] \over (x^2-4)^3 }$

= 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 $x \approx 5.70$ . 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 ($\uparrow$) for x<-2 , -2<x<1 , and x>4 ;

f is ( $\downarrow$) 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 ($\cup$) for x<-2 and 2<x<5.70 ;

f is ($\cap$) for -2<x<2 and x>5.70 ;

f has an inflection point at $x \approx 5.70$ , $y \approx 0.10$ .

OTHER INFORMATION ABOUT f :

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

$${ \lim_{ x \to \pm \infty } \ { (x-4)^2 \over x^2-4 } } = \disp... ...over x^2-4 } { \displaystyle{1 \over x^2} \over \displaystyle{1 \over x^2} } }$$

$= \displaystyle{ \lim_{ x \to \pm \infty } \ { 1 - \displaystyle{8 \over x } + \displaystyle{16 \over x^2 } \over 1 - \displaystyle{ 4 \over x^2 } } }$

$= \displaystyle{ 1 - 0 + 0 \over 1 - 0 }$

= 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,

$${ \lim_{ x \to 2^{+} } \ { (x-4)^2 \over x^2-4 } } = { (-2)^2 \over (2)^2 - 4 }$$

$= \displaystyle{ 4 \over 0^{+} }$

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

$= + \infty$ ,

and

$${ \lim_{ x \to 2^{-} } \ { (x-4)^2 \over x^2-4 } } = { (-2)^2 \over (2)^2 - 4 }$$

$= \displaystyle{ 4 \over 0^{-} }$

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

$= -\infty$ .

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 $\infty$ or $-\infty$, a vertical asymptote exists. Now check at x=-2 . Thus,

$${ \lim_{ x \to -2^{+} } \ { (x-4)^2 \over x^2-4 } } = { (-6)^2 \over (-2)^2 - 4 }$$

$= \displaystyle{ 36 \over 0^{-} }$

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

$= -\infty$ ,

and

$${ \lim_{ x \to -2^{-} } \ { (x-4)^2 \over x^2-4 } } = { (-6)^2 \over (-2)^2 - 4 }$$

$= \displaystyle{ 36 \over 0^{+} }$

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

$= + \infty$ .

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

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