Curve Sketching 1

Years

Year 11

Derivative Curve Sketching

Curve Sketching

Curve Sketching 1

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Question 1 of 5

1. Question
Sketch the First Derivative Curve of the curve shown below:
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A point that satisfies $f'(x)=0$ is called a Stationary Point

A positive $f'(x)$ means a positive slope

A negative $f'(x)$ means a negative slope

First, set up a separate plane with the $y$ axis replaced with $f'(x)$ axis

Next, mark out the stationary points from the given curve into the revised graph

Note that in the revised graph, stationary points are always on the $x$ axis

For an easier transfer, mark out the movement of the given curve

Remember that a positive or increasing slope means $f'(x)$ is positive, and a negative or decreasing slope means $f'(x)$ is negative

Finally, draw a curve on the revised graph depending on the signs marked on the original curve
Question 2 of 5

2. Question
Sketch the curve:

$f(x)=x^2-2x$
- 1.
- 2.
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A point that satisfies $f'(x)=0$ is called a Stationary Point

A positive $f'(x)$ means a positive slope

A negative $f'(x)$ means a negative slope

First, equate $f'(x)$ to $0$ to find the stationary points

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

$f'(x)$ $=$ $2x-2$

$2x-2$ $=$ $0$ Equate to $0$

$2(x-1)$ $=$ $0$ Factor out $2$

$x$ $=$ $1$

This means that there is a stationary point at $x=1$

Find the corresponding $y$ value of $x=1$

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

$f(\color{#007DDC}{1})$ $=$ $\color{#007DDC}{1}^2)-2(\color{#007DDC}{1})$

$=$ $1-2$

$=$ $-1$

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

Next, graph $f'(x)=2x-2$ to identify where the curve increases or decreases

Start by setting up a plane with the $y$ axis replaced with $f'(x)$ axis and plot $x=1$

Substitute $x=0$ to know where the derivative graph intersects the $f'(x)$ axis

$f'(x)$ $=$ $2x-2$

$=$ $2(0)-2$ Substitute $x=0$

$=$ $0-2$

$f'(x)$ $=$ $-2$

Form a line by connecting the two points

From the diagram above, we can say the following about the curve of $f(x)$ :

The slope is negative or decreasing at $x≤1$

The slope is positive or increasing at $x≥1$

Next, check where the curve intersects the $x$ axis by equating $f(x)$ to $0$

$f(x)$ $=$ $0$

$x^2-2x$ $=$ $0$ Equate to $0$

$x(x-2)$ $=$ $0$

$x$ $=$ $0,2$

The curve intersects $x=0$ and $x=2$

Lastly, draw a curve or a parabola over the plotted points, following the increase and decrease that have been set with the derivative graph

Question 3 of 5

Sketch the curve

$f(x)=x^4-8x^2+16$

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A point that satisfies

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

A sign diagram is a way of visualizing a curve’s graph by indicating the inflection points, stationary points, and the increases or decreases in the curve.

First, equate

$f'(x)$ to

$0$ to find the stationary points

$f(x)$	$=$	$x^4-8x^2+16$
$f'(x)$	$=$	$4x^3-16x=0$	Equate $f'(x)$ to $0$
	$=$	$4x(x^2-4)=0$	Factor out $4x$
	$=$	$4x(x-2)(x+2)=0$	Factor out the difference of $2$ squares
		$x=0,x=2,x=-2$

Find the corresponding

$y$ values by substituting each

$x$ value to the function

$f(x)$	$=$	$x^4-8x^2+16$
$f(\color{#007DDC}{0})$	$=$	$\color{#007DDC}{0}^4-8(\color{#007DDC}{0}^2)+16$
	$=$	$0-0+16$
	$=$	$16$

Stationary Point:

$(\color{#007DDC}{0},\color{#e65021}{16})$

$f(x)$	$=$	$x^4-8x^2+16$
$f(\color{#007DDC}{2})$	$=$	$\color{#007DDC}{2}^4-8(\color{#007DDC}{2}^2)+16$
	$=$	$16-32+16$
	$=$	$0$

Stationary Point:

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

$f(x)$	$=$	$x^4-8x^2+16$
$f(\color{#007DDC}{-2})$	$=$	$\color{#007DDC}{-2}^4-8(\color{#007DDC}{-2}^2)+16$
	$=$	$16-32+16$
	$=$	$0$

Stationary Point:

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

Next, create a sign diagram to identify where the curve increases or decreases

Start by setting up a horizontal line with matching indicators of the stationary points

Test the gradient of a point to the left of

$-2$ such as $x=-3$

$f'(x)$	$=$	$4x^3-16x$
$f'(\color{#00880A}{-3})$	$=$	$4(\color{#00880A}{-3})^3-16(\color{#00880A}{-3})$
	$=$	$4(-27)+48$
	$=$	$-108+48$
	$=$	$-60$

This value is negative, which means the curve’s slope at $-3$ is decreasing

Indicate this on the sign diagram by adding a negative sign to the left of

$-2$

Test the gradient of a point to the right of

$-2$ such as $x=-1$

$f'(x)$	$=$	$4x^3-16x$
$f'(\color{#00880A}{-1})$	$=$	$4(\color{#00880A}{-1})^3-16(\color{#00880A}{-1})$
	$=$	$4(-1)+16$
	$=$	$-4+16$
	$=$	$12$

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

Indicate this on the sign diagram by adding a positive sign to the right of

$-2$

Test the gradient of a point to the left of

$2$ such as $x=1$

$f'(x)$	$=$	$4x^3-16x$
$f'(\color{#00880A}{1})$	$=$	$4(\color{#00880A}{1}^3)-16(\color{#00880A}{1})$
	$=$	$4(1)-16$
	$=$	$4-16$
	$=$	$-12$

This value is negative, which means the curve’s slope at $1$ is decreasing

Indicate this on the sign diagram by adding a negative sign to the left of

$2$

Test the gradient of a point to the right of

$2$ such as $x=3$

$f'(x)$	$=$	$4x^3-16x$
$f'(\color{#00880A}{3})$	$=$	$4(\color{#00880A}{3}^3)-16(\color{#00880A}{3})$
	$=$	$4(27)-48$
	$=$	$108-48$
	$=$	$60$

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

Indicate this on the sign diagram by adding a positive sign to the left of

$2$

Finally, draw a curve along the stationary points with the help of the sign diagram

Question 4 of 5

To help identify where the curve increases or decreases, sketch the first derivative of the curve:

$f(x)=x^3-6x^2+9x+1$

Question 5 of 5

Sketch the curve:

$f(x)=x^3-6x^2+9x+1$

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A point that satisfies

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

A positive $f'(x)$ means a positive slope

A negative $f'(x)$ means a negative slope

First, equate

$f'(x)$ to

$0$ to find the stationary points

$f(x)$	$=$	$x^3-6x^2+9x+1$
$f'(x)$	$=$	$3x^2-12x+9$
$3x^2-12x+9$	$=$	$0$	Equate to $0$
$3(x^2-4x+3)$	$=$	$0$	Factor out $3$
$3(x-3)(x-1)$	$=$	$0$	Factor out further
$x$	$=$	$3,1$

This means that there are stationary points at $x=3$ and $x=1$

Find the corresponding

$y$ values by substituting each

$x$ value to the function

$f(x)$	$=$	$x^3-6x^2+9x+1$
$f(\color{#007DDC}{3})$	$=$	$\color{#007DDC}{3}^3-6(\color{#007DDC}{3}^2)+9(\color{#007DDC}{3})+1$
	$=$	$27-54+27+1$
	$=$	$1$

Stationary Point:

$(\color{#007DDC}{3},\color{#e65021}{1})$

$f(x)$	$=$	$x^3-6x^2+9x+1$
$f(\color{#007DDC}{1})$	$=$	$\color{#007DDC}{1}^3-6(\color{#007DDC}{1}^2)+9(\color{#007DDC}{1})+1$
	$=$	$1-6+9+1$
	$=$	$5$

Stationary Point:

$(\color{#007DDC}{1},\color{#e65021}{5})$

Next, graph

$f'(x)=3x^2-12x+9$ to identify where the curve increases or decreases

Start by setting up a plane with the

$y$ axis replaced with

$f'(x)$ axis and plot $x=1$ and $x=3$

Since the highest power of

$f'(x)$ is

$2$ , draw a parabola over the plotted points

From the diagram above, we can say the following about the curve of

$f(x)$ :

The slope is positive or increasing at

$x≤1$

The slope is negative or decreasing at

$1≤x≤3$

The slope is positive or increasing at

$x≥3$

Next, check where the curve intersects the

$y$ axis by substituting

$x=0$ to

$f(x)$

$f(x)$	$=$	$x^3-6x^2+9x+1$
$f(0)$	$=$	$0^3-6(0^2)+9(0)+1$	Substitute $x=0$
	$=$	$0-0+0+1$
$y$	$=$	$1$

The curve intersects

$y=1$

Lastly, draw a curve over the plotted points, following the increase and decrease that have been set with the derivative graph

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Quiz summary

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1. Question

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2. Question

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3. Question

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4. Question

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5. Question

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Quizzes

$f(x)$	$=$	$x^2-2x$
$f'(x)$	$=$	$2x-2$
$2x-2$	$=$	$0$	Equate to $0$
$2(x-1)$	$=$	$0$	Factor out $2$
$x$	$=$	$1$

$f'(x)$	$=$	$2x-2$
	$=$	$2(0)-2$	Substitute $x=0$
	$=$	$0-2$
$f'(x)$	$=$	$-2$