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) x^1/3-1

= 1 - x^-2/3

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

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

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

= 0

when x^2/3 - 1 = 0 . Thus, x^2/3 = 1 so that $x^2 = \big( x^{2/3} \big)^3 = 1^3 = 1$ 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

$= \displaystyle{ 2 \over 3 x^{5/3} }$

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

f is ( $\downarrow$) 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 ($\cup$) for x>0 ;

f is ($\cap$) 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 - 3x^1/3 = x^1/3 ( x^2/3 - 3 ) = 0 so that x^1/3=0 or x^2/3 = 3 . Thus, $x= \big( x^{1/3} \big)^3 = (0)^3 = 0$ or $x^2 = \big( x^{2/3} \big)^{3} = (3)^3 = 27$ , so that $x= \pm \sqrt{27} = \pm 3\sqrt{3} \approx \pm 5.20$ 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

$f(x) = x^{2/3} \Big( 5/2 - x \Big)$

= (5/2)x^2/3 - x^2/3 x¹

= (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

$= \displaystyle{ 5 \over 3 x^{1/3} } - \displaystyle{ 5 x^{2/3} \over 3 }$

$= \displaystyle{ 5 \over 3 x^{1/3} } - \displaystyle{ 5 x^{2/3} \over 3 } \displaystyle{ x^{1/3} \over x^{1/3} }$

$= \displaystyle{ 5 \over 3 x^{1/3} } - \displaystyle{ 5 x^{2/3+1/3} \over 3x^{1/3} }$

$= \displaystyle{ 5 - 5x \over 3 x^{1/3} }$

$= \displaystyle{ 5(1- x) \over 3 x^{1/3} }$

= 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

$= \displaystyle{ -5 \over 9 x^{4/3} } - \displaystyle{ 10 \over 9 x^{1/3} }$

$= \displaystyle{ -5 \over 9 x^{4/3} } - \displaystyle{ 10 \over 9 x^{1/3} } \displaystyle{ x \over x }$

$= \displaystyle{ -5 \over 9 x^{4/3} } - \displaystyle{ 10x \over 9 x^{1/3} x^1 }$

$= \displaystyle{ -5 \over 9 x^{4/3} } - \displaystyle{ 10x \over 9 x^{1/3+1} }$

$= \displaystyle{ -5 \over 9 x^{4/3} } - \displaystyle{ 10x \over 9 x^{4/3} }$

$= \displaystyle{ -5 -10x \over 9 x^{4/3} }$

$= \displaystyle{ -5(1+2x) \over 9 x^{4/3} }$

= 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 ($\uparrow$) for 0<x<1 ;

f is ( $\downarrow$) 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 ($\cup$) for x< -1/2 ;

f is ($\cap$) for -1/2<x<0 and x>0 ;

f has an inflection point at x=-1/2 , $y=3(1/2)^{2/3} \approx 1.89$ .

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 $0 \le x \le 2\pi$ . Now determine a sign chart for the first derivative, f' . Begin with

$f'(x) = \cos x - \sqrt{3} (-\sin x)$

$= \cos x + \sqrt{3} \sin x$

= 0 .

Then

$\sqrt{3} \sin x = -\cos x$

so that

$${ \sin x \over \cos x } = \displaystyle{ -1 \over \sqrt{3} } = \displaystyle{ -1/2 \over \sqrt{3}/2 }$$

or

$${ \sin x \over \cos x } = \displaystyle{ -1 \over \sqrt{3} } = \displaystyle{ 1/2 \over -\sqrt{3}/2 }$$ .

A well-known angle x in the second quadrant with

$\cos x = -\sqrt{3}/2$ and $\sin x = 1/2$

is $x= 5\pi/6$ . A well-known angle x in the fourth quadrant with

$\cos x = \sqrt{3}/2$ and $\sin x = -1/2$

is $x= 11\pi/6$ . It is convenient to use $x=0, \pi ,$ and $2\pi$ as test points in the first derivative $f'(x) = \cos x + \sqrt{3} \sin x$ when constructing the adjoining sign chart for the first derivative, f' .

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

$f''(x) = -\sin x + \sqrt{3} \cos x$

= 0 .

Then

$\sin x = \sqrt{3} \cos x$

so that

$${ \sin x \over \cos x } = \sqrt{3} = \displaystyle{ \sqrt{3}/2 \over 1/2 }$$

or

$${ \sin x \over \cos x } = \sqrt{3} = \displaystyle{ -\sqrt{3}/2 \over -1/2 }$$ .

A well-known angle x in the first quadrant with

$\cos x = 1/2$ and $\sin x = \sqrt{3}/2$

is $x= \pi/3$ . A well-known angle x in the third quadrant with

$\cos x = -1/2$ and $\sin x = -\sqrt{3}/2$

is $x= 4\pi/3$ . It is convenient to use $x=0, \pi ,$ and $2\pi$ as test points in the second derivative $f''(x) = -\sin x + \sqrt{3} \cos x$ when constructing the adjoining sign chart for the second derivative, f'' .

Now summarize the information from each sign chart.

FROM f' :

f is ($\uparrow$) for $0<x<5\pi/6$ and $11\pi/6<x<2\pi$ ;

f is ( $\downarrow$) for $5\pi/6<x<11\pi/6$ ;

f has a relative maximum at $x=2\pi$ , $y= -\sqrt{3}$ ;

f has an absolute maximum at $x= 5\pi/6$ , y= 2 ;

f has a relative minimum at x=0 , $y= -\sqrt{3}$

f has an absolute minimum at $x= 11\pi/6$ , y=-2 .

FROM f'' :

f is ($\cup$) for $0<x<\pi/3$ and $4\pi/3<x<2\pi$ ;

f is ($\cap$) for $\pi/3<x<4\pi/3$ ;

f has inflection points at $x= \pi/3$ , y=0 and $x= 4\pi/3$ , y=0 .

OTHER INFORMATION ABOUT f :

If x=0 , then $y= -\sqrt{3}$ , so that $y= -\sqrt{3}$ is the y-intercept. If y=0 , then $\sin x - \sqrt{ 3 } \cos x=0$ and $\sin x = \sqrt{3} \cos x$ (This is the same equation used to solve f''(x)=0 .) so that $x= \pi/3$ and $x= 4\pi/3$ 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 $f(x) = x \sqrt{ 4 - x^2 } = x \sqrt{ (2-x)(2+x) }$ the domain of f is $-2 \le x \le 2$ 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

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

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

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

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

$= \displaystyle{ 4-2x^2 \over \sqrt{4-x^2} }$

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

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

= 0

for x= 0 , $x = \sqrt{2}$ ,and $x=-\sqrt{2}$ . 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

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

and using the quotient rule, we get

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

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

$= { { \displaystyle{ -4x \sqrt{4-x^2} \over 1 } \displaystyle{ \sqrt{4-x^2} \o... ...style{ x(4-2x^2) \over \sqrt{4-x^2} } } \over \displaystyle{ 4-x^2 \over 1 } }$

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

(Factor out 2x .)

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

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

$= \displaystyle{ 2x(x^2-6) \over (4-x^2)^{1+1/2} }$

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

= 0 .

Then $2x(x-\sqrt{6})(x+\sqrt{6})=0$ so that x= 0 , $x= \sqrt{6}$ , or $x=-\sqrt{6}$ . HOWEVER, $x= \sqrt{6}$ and $x=-\sqrt{6}$ 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 ($\uparrow$) for $-\sqrt{2}<x<\sqrt{2}$ ;

f is ( $\downarrow$) for $-2<x<-\sqrt{2}$ and $\sqrt{2}<x<2$ ;

f has an absolute maximum at $x = \sqrt{2}$ , $y=2\sqrt{2}$ ;

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

f has an absolute minimum at $x=-\sqrt{2}$ , $y=-2\sqrt{2}$ ;

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

FROM f'' :

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

f is ($\cap$) 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 $x \sqrt{ 4 - x^2 } = x \sqrt{ (2-x)(2+x) }=0$ 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 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.

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