Curve Sketching 2

Question 1 of 5

Sketch the curve

$f(x)=3x^4-16x^3+24x^2+3$

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A point that satisfies

$f'(x)=0$ is called a Stationary Point

A point that satisfies

$f”(x)=0$ is called an Inflection Point

First, equate

$f'(x)$ to

$0$ to find the stationary points

$f(x)$	$=$	$3x^4-16x^3+24x^2+3$
$f'(x)$	$=$	$12x^3-48x^2+48x=0$	Equate $f'(x)$ to $0$
	$=$	$12x(x^2-4x+4)=0$	Factor out $12x$
	$=$	$4x(x-2)(x-2)=0$	Factor out further
		$x=0,x=2$

Find the corresponding

$y$ values by substituting each

$x$ value to the function

$f(x)$	$=$	$3x^4-16x^3+24x^2+3$
$f(\color{#007DDC}{0})$	$=$	$3(\color{#007DDC}{0}^4)-16(\color{#007DDC}{0}^3)+24(\color{#007DDC}{0}^2)+3$
	$=$	$0-0+0+3$
	$=$	$3$

Stationary Point:

$(\color{#007DDC}{0},\color{#e65021}{3})$

$f(x)$	$=$	$3x^4-16x^3+24x^2+3$
$f(\color{#007DDC}{2})$	$=$	$3(\color{#007DDC}{2}^4)-16(\color{#007DDC}{2}^3)+24(\color{#007DDC}{2}^2)+3$
	$=$	$48-128+96+3$
	$=$	$19$

Stationary Point:

$(\color{#007DDC}{2},\color{#e65021}{19})$

Use the second derivative to check if each stationary point is either a maximum, minimum, or an inflection point

Checking

$(\color{#007DDC}{0},\color{#e65021}{3})$ :

$f'(x)$	$=$	$12x^3-48x^2+48x$
$f”(x)$	$=$	$36x^2-96x+48$
$f”(\color{#007DDC}{0})$	$=$	$36(\color{#007DDC}{0}^2)-96(\color{#007DDC}{0})+48$
	$=$	$0-0+48$
	$=$	$48$ $>$ $0$

This means

$(\color{#007DDC}{0},\color{#e65021}{3})$ is a minimum point

Checking

$(\color{#007DDC}{2},\color{#e65021}{19})$ :

$f'(x)$	$=$	$12x^3-48x^2+48x$
$f”(x)$	$=$	$36x^2-96x+48$
$f”(\color{#007DDC}{2})$	$=$	$36(\color{#007DDC}{2}^2)-96(\color{#007DDC}{2})+48$
	$=$	$144-192+48$
	$=$	$0$

This means

$(\color{#007DDC}{2},\color{#e65021}{19})$ is an inflection point

Identify where the curve decreases or increases

It is clear that a minimum point would have a decreasing curve on the left and an increasing curve on the right. Hence, we just check the curves around the inflection point

Start by setting up a grid of points around $x=2$

Test the gradient at $x=1$

$f'(x)$	$=$	$12x^3-48x^2+48x=0$
$f'(\color{#00880A}{1})$	$=$	$12(\color{#00880A}{1})^3-48(\color{#00880A}{1}^2)+48(\color{#00880A}{1})$
	$=$	$12(1)-48(1)+48$
	$=$	$12-48+48$
	$=$	$12$

This value is positive, which means the curve’s slope at $1$ is increasing

Indicate this by adding an increasing slant line under $1$

Test the gradient at $x=3$

$f'(x)$	$=$	$12x^3-48x^2+48x=0$
$f'(\color{#00880A}{3})$	$=$	$12(\color{#00880A}{3})^3-48(\color{#00880A}{3}^2)+48(\color{#00880A}{3})$
	$=$	$12(27)-48(9)+144$
	$=$	$324-432+144$
	$=$	$36$

This value is positive, which means the curve’s slope at $3$ is increasing

Indicate this by adding an increasing slant line under $3$

Finally, draw a curve along the stationary points, keeping in mind where it decreases or increases

Question 2 of 5

Sketch the curve

$f(x)=(x^2-1)/(x^2+1)$

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A point that satisfies

$f'(x)=0$ is called a Stationary Point

A point that satisfies

$f”(x)=0$ is called an Inflection Point

First, equate

$f'(x)$ to

$0$ to find the stationary points

$f(x)$	$=$	$(x^2-1)/(x^2+1)$
$f'(x)$	$=$	$(4x)/((x^2+1)^2)=0$	Equate $f'(x)$ to $0$
	$=$	$4x=0$	Multiply both sides by $(x^2+1)^2$
	$=$	$0=0$	$x=0$ will satisfy the equation
		$x=0$

Find the corresponding

$y$ values by substituting each

$x$ value to the function

$f(x)$	$=$	$(x^2-1)/(x^2+1)$
$f(\color{#007DDC}{0})$	$=$	$\frac{\color{#007DDC}{0}^2-1}{\color{#007DDC}{0}^2+1}$
	$=$	$-1/1$
	$=$	$-1$

Stationary Point:

$(\color{#007DDC}{0},\color{#e65021}{-1})$

Use the second derivative to check if the stationary point

$($ $0$

$,$ $-1$

$)$ is either a maximum, minimum, or an inflection point

$f'(x)$	$=$	$(4x)/((x^2+1)^2)$
$f”(x)$	$=$	$(4(x^2+1)^2-16x(x^2+1)^2)/((x^2+1)^4)$
$f”(\color{#007DDC}{0})$	$=$	$\frac{4(\color{#007DDC}{0}^2+1)^2-16(\color{#007DDC}{0})(\color{#007DDC}{0}^2+1)^2}{(\color{#007DDC}{0}^2+1)^4}$
	$=$	$4$ $>$ $0$

This means

$(\color{#007DDC}{0},\color{#e65021}{-1})$ is a minimum point

Next, solve for the

$y$ and

$x-$ intercepts of the curve by substituting

$x=0$ and

$y=0$ respectively to the formula

$y-$ intercept

$y$	$=$	$(x^2-1)/(x^2+1)$

$y$	$=$	$(0^2-1)/(0^2+1)$	Substitute $x=0$

$y$	$=$	$-1/1$

$y$	$=$	$-1$

$x-$ intercept

$y$	$=$	$(x^2-1)/(x^2+1)$

$0$	$=$	$(x^2-1)/(x^2+1)$	Substitute $y=0$

$0$	$=$	$x^2-1$	Multiply both sides by $x^2+1$
$1$	$=$	$x^2$	Add $1$ to both sides
$x$	$=$	$+-1$	Find the square root of both sides

Next, apply the limit

$∞$ to

$x$ to check for asymptotes

$lim_(x→∞)(x^2-1)/(x^2+1)$

$=$

$1$

This means that the curve will be approaching but will not go beyond

$y=1$

Finally, draw a curve along the stationary points and intercepts, keeping in mind not to touch the asymptote

Question 3 of 5

Sketch the curve

$f(x)=(x-4)(x+2)^2$

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A point that satisfies

$f'(x)=0$ is called a Stationary Point

A point that satisfies

$f”(x)=0$ is called an Inflection Point

First, equate

$f'(x)$ to

$0$ to find the stationary points

$f(x)$	$=$	$(x-4)(x+2)^2$
$f'(x)$	$=$	$3x^3-12=0$	Equate $f'(x)$ to $0$
	$=$	$3(x^2-4)=0$	Factor out $3$
	$=$	$3(x+2)(x-2)=0$	Factor out further
		$x=-2,x=2$

Find the corresponding

$y$ values by substituting each

$x$ value to the function

$f(x)$	$=$	$(x-4)(x+2)^2$
$f(\color{#007DDC}{-2})$	$=$	$[(\color{#007DDC}{-2})-4][(\color{#007DDC}{-2})+2]^2$
	$=$	$(-6)(0)$
	$=$	$0$

Stationary Point:

$(\color{#007DDC}{-2},\color{#e65021}{0})$

$f(x)$	$=$	$(x-4)(x+2)^2$
$f(\color{#007DDC}{2})$	$=$	$(\color{#007DDC}{2}-4)(\color{#007DDC}{2}+2)^2$
	$=$	$(-2)(16)$
	$=$	$-32$

Stationary Point:

$(\color{#007DDC}{2},\color{#e65021}{-32})$

Use the second derivative to check if each stationary point is either a maximum, minimum, or an inflection point

Checking

$(\color{#007DDC}{-2},\color{#e65021}{0})$ :

$f'(x)$	$=$	$3x^3-12$
$f”(x)$	$=$	$6x$
$f”(\color{#007DDC}{-2})$	$=$	$6(\color{#007DDC}{-2})$
	$=$	$-12$ $<$ $0$

This means

$(\color{#007DDC}{-2},\color{#e65021}{0})$ is a maximum point

Checking

$(\color{#007DDC}{2},\color{#e65021}{-32})$ :

$f'(x)$	$=$	$12x^3-48x^2+48x$
$f”(x)$	$=$	$3x^3-12$
$f”(\color{#007DDC}{2})$	$=$	$6(\color{#007DDC}{2})$
	$=$	$12$ $>$ $0$

This means

$(\color{#007DDC}{2},\color{#e65021}{-32})$ is a minimum point

Next, substitute

$x=0$ to the original function to find the inflection point

$f(x)$	$=$	$(x-4)(x+2)^2$
$f(0)$	$=$	$(0-4)(0+2)^2$	Equate $x=0$
	$=$	$(-4)(4)$
	$=$	$-16$

This means

$(0,-16)$ is an inflection point

Finally, draw a curve along the stationary points, keeping in mind where it decreases or increases

Question 4 of 5

Sketch the curve

$f(x)=x+1/x$

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A point that satisfies

$f'(x)=0$ is called a Stationary Point

A point that satisfies

$f”(x)=0$ is called an Inflection Point

First, equate

$f'(x)$ to

$0$ to find the stationary points

$f(x)$	$=$	$x+1/x$
$f'(x)$	$=$	$1-1/(x^2)=0$	Equate $f'(x)$ to $0$
	$=$	$x^2-1=0$	Multiply both sides by $x^2$
	$=$	$x^2=1$	Factor the difference of two squares
		$x=1,x=-1$

Find the corresponding

$y$ values by substituting each

$x$ value to the function

$f(x)$	$=$	$x+1/x$
$f(\color{#007DDC}{1})$	$=$	$\color{#007DDC}{1}+\frac{1}{\color{#007DDC}{1}}$
	$=$	$1+1$
	$=$	$2$

Stationary Point:

$(\color{#007DDC}{1},\color{#e65021}{2})$

$f(x)$	$=$	$x+1/x$
$f(\color{#007DDC}{-1})$	$=$	$\color{#007DDC}{-1}+\frac{1}{\color{#007DDC}{-1}}$
	$=$	$-1-1$
	$=$	$-2$

Stationary Point:

$(\color{#007DDC}{-1},\color{#e65021}{-2})$

Use the second derivative to check if each stationary point is either a maximum, minimum, or an inflection point

Checking

$(\color{#007DDC}{1},\color{#e65021}{2})$ :

$f'(x)$	$=$	$1-1/(x^2)$	$=$	$1-x^(-2)$
$f”(x)$	$=$	$2x^(-3)$	$=$	$2/(x^3)$
$f”(\color{#007DDC}{1})$	$=$	$\frac{2}{\color{#007DDC}{1}^3}$
	$=$	$2$ $>$ $0$

This means

$(\color{#007DDC}{1},\color{#e65021}{2})$ is a minimum point

Checking

$(\color{#007DDC}{-1},\color{#e65021}{-2})$ :

$f'(x)$	$=$	$1-1/(x^2)$	$=$	$1-x^(-2)$
$f”(x)$	$=$	$2x^(-3)$	$=$	$2/(x^3)$
$f”(\color{#007DDC}{-1})$	$=$	$\frac{2}{\color{#007DDC}{-1}^3}$
	$=$	$-2$ $<$ $0$

This means

$(\color{#007DDC}{-1},\color{#e65021}{-2})$ is a maximum point

Next, apply the limits

$0$ and

$∞$ to

$x$ to check for asymptotes

$x→0$

$lim_(x→0)1+1/x$

$=$

$\text(undefined)$

$x→∞$

$lim_(x→∞)1+1/x$

$=$

$\text(undefined)$

Finally, draw a curve along the stationary points and between but not meeting the asymptotes

Question 5 of 5

Graph the equation

$y=(x-1)e^x$

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Product Rule

$y’=vu’+uv’$

First, solve for the first derivative using the product rule

$u$	$=$	$x-1$
$v$	$=$	$e^x$
$u’$	$=$	$1$
$v’$	$=$	$e^x$

$y’$	$=$	$vu’+uv’$
	$=$	$e^x(1)+(x-1)e^x$	Substitute known values
	$=$	$e^x+(x-1)e^x$	Evaluate
	$=$	$e^x+xe^x-e^x$	Distribute
	$=$	$xe^x$	$e^x-e^x$ cancels out

Next, equate the first derivative to

$0$ to find the stationary point

$y’=xe^x$	$=$	$0$
$xe^x$ $divide e^x$	$=$	$0$ $divide e^x$	Divide both sides by $e^x$
$x$	$=$	$0$

To get the value of

$y$ , substitute the value of

$x$ to the equation

$y$	$=$	$(x-1)e^x$
	$=$	$(\color{#CC0000}{0}-1)e^{\color{#CC0000}{0}}$	$x=0$
	$=$	$(-1)1$	Anything raised to $0$ is $1$
	$=$	$-1$

Therefore, the stationary point will be

$(0,-1)$

Next, determine if the stationary point would be a concave up or concave down

To do this, use the product rule on the first derivative and get the second derivative

$u$	$=$	$x$
$v$	$=$	$e^x$
$u’$	$=$	$1$
$v’$	$=$	$e^x$

$y”$	$=$	$vu’+uv’$
	$=$	$e^x(1)+(x)e^x$	Substitute known values
	$=$	$e^x+xe^x$	Evaluate
	$=$	$e^x(1+x)$	Factorize

Substitute the value of

$x$ to the second derivative to determine the concavity

$y”$	$=$	$e^x(1+x)$
	$=$	$e^{\color{#CC0000}{0}}(1+\color{#CC0000}{0})$	$x=0$
	$=$	$1times1$	Anything raised to $0$ is $1$
	$=$	$1$

A positive concavity means that the stationary point will be a minimum, or the concave will be upward

Next, equate the second derivative to

$0$ to find the inflexion points

$y’=e^x(1+x)$	$=$	$0$
$e^x(1+x)$ $divide e^x$	$=$	$0$ $divide e^x$	Divide both sides by $e^x$
$1+x$ $-1$	$=$	$0$ $-1$	Subtract $1$ from both sides
$x$	$=$	$-1$

Substitute this value of

$x$ to the equation to find the value of

$y$

$y$	$=$	$(x-1)e^x$
	$=$	$(\color{#CC0000}{-1}-1)e^{\color{#CC0000}{-1}}$	$x=0$

	$=$	$-2/e$	Reciprocate $e^(-1)$

Therefore, the inflexion point will be

$(0,-2/e)$

$x=1$ is the only x-intercept, hence the curve will not be touching the x-axis on the left side

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Product Rule

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