Math / Calculus

Question12. [12 marks] Consider the function $y=-x^{4}+4 x^{3}$ . a. Find the $x$ -and $y$ -intercepts of the function. $x-i n t=(4,0), y-i n t:(0,0)$ b. Determine the intervals where the function is increasing/decreasing and find the relative extrema. c. Determine the intervals where the function is concave up/ concave down and find the inflection points. d. Sketch the function.

Studdy Solution

STEP 1

What is this asking? We're going to explore this cool function $y = -x^4 + 4x^3$ , find its intercepts, figure out where it's increasing and decreasing, where it's curving up or down, and then draw a sweet sketch of it! Watch out! Don't mix up increasing/decreasing with concave up/concave down.
They're related, but totally different things!
Also, remember that relative extrema are like peaks and valleys, and inflection points are where the curve changes its bendiness.

STEP 2

1. Find the intercepts
2. Analyze increasing/decreasing behavior and extrema
3. Analyze concavity and inflection points
4. Sketch the graph

STEP 3

To find the y-intercept, we set $x = 0$ in our function: $y = -(0)^4 + 4(0)^3 = 0$ .
So, the y-intercept is at the point $(0, 0)$ .
Boom!

STEP 4

For the x-intercepts, we set $y = 0$ and solve for $x$ : $0 = -x^4 + 4x^3$ $0 = x^3(-x + 4)$ This equation is true if $x^3 = 0$ or if $-x + 4 = 0$ .
So, our x-intercepts are $x = 0$ and $x = 4$ , giving us the points $(0, 0)$ and $(4, 0)$ .
Awesome!

STEP 5

First, we need the first derivative of our function, which tells us about the slope: $y' = -4x^3 + 12x^2$

STEP 6

To find where the function is increasing or decreasing, we set the first derivative equal to zero and solve for $x$ : $0 = -4x^3 + 12x^2$ $0 = -4x^2(x - 3)$ So, $x = 0$ and $x = 3$ are our critical points.

STEP 7

Now, we test values around these critical points to see if the derivative is positive (increasing) or negative (decreasing).
Let's pick $x = -1$ , $x = 1$ , and $x = 4$ . * For $x = -1$ , $y' = -4(-1)^3 + 12(-1)^2 = 4 + 12 = 16 > 0$ , so the function is increasing before $x = 0$ . * For $x = 1$ , $y' = -4(1)^3 + 12(1)^2 = -4 + 12 = 8 > 0$ , so the function is increasing between $x = 0$ and $x = 3$ . * For $x = 4$ , $y' = -4(4)^3 + 12(4)^2 = -256 + 192 = -64 < 0$ , so the function is decreasing after $x = 3$ .

STEP 8

Since the function goes from increasing to decreasing at $x = 3$ , we have a relative maximum there.
Let's find the y-value: $y = -(3)^4 + 4(3)^3 = -81 + 108 = 27$ .
So, the relative maximum is at $(3, 27)$ .

STEP 9

We need the second derivative to talk about concavity: $y'' = -12x^2 + 24x$

STEP 10

We set the second derivative equal to zero and solve for $x$ : $0 = -12x^2 + 24x$ $0 = -12x(x - 2)$ So, $x = 0$ and $x = 2$ are our potential inflection points.

STEP 11

Let's test $x = -1$ , $x = 1$ , and $x = 3$ : * For $x = -1$ , $y'' = -12(-1)^2 + 24(-1) = -12 - 24 = -36 < 0$ , so concave down. * For $x = 1$ , $y'' = -12(1)^2 + 24(1) = -12 + 24 = 12 > 0$ , so concave up. * For $x = 3$ , $y'' = -12(3)^2 + 24(3) = -108 + 72 = -36 < 0$ , so concave down.

STEP 12

Since the concavity changes at both $x = 0$ and $x = 2$ , these are inflection points.
We already know the y-value for $x = 0$ is $0$ .
For $x = 2$ , $y = -(2)^4 + 4(2)^3 = -16 + 32 = 16$ .
So, our inflection points are $(0, 0)$ and $(2, 16)$ .

STEP 13

Using all this info, we can sketch the graph!
It starts increasing, passes through $(0,0)$ , keeps increasing and changes from concave down to concave up at $(0,0)$ , reaches an inflection point at $(2,16)$ , hits a maximum at $(3,27)$ , then decreases, passing through $(4,0)$ , and continues decreasing while being concave down.

STEP 14

The x-intercepts are $(0, 0)$ and $(4, 0)$ , and the y-intercept is $(0, 0)$ .
The function is increasing on $(-\infty, 3)$ and decreasing on $(3, \infty)$ , with a relative maximum at $(3, 27)$ .
It's concave up on $(0, 2)$ and concave down on $(-\infty, 0)$ and $(2, \infty)$ , with inflection points at $(0, 0)$ and $(2, 16)$ .

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Studdy Solution

STEP 1

STEP 2

1. Find the intercepts
2. Analyze increasing/decreasing behavior and extrema
3. Analyze concavity and inflection points
4. Sketch the graph

STEP 3

To find the y-intercept, we set $x = 0$ in our function: $y = -(0)^4 + 4(0)^3 = 0$ .
So, the y-intercept is at the point $(0, 0)$ .
Boom!

STEP 4

For the x-intercepts, we set $y = 0$ and solve for $x$ : $0 = -x^4 + 4x^3$ $0 = x^3(-x + 4)$ This equation is true if $x^3 = 0$ or if $-x + 4 = 0$ .
So, our x-intercepts are $x = 0$ and $x = 4$ , giving us the points $(0, 0)$ and $(4, 0)$ .
Awesome!

STEP 5

First, we need the first derivative of our function, which tells us about the slope: $y' = -4x^3 + 12x^2$

STEP 6

To find where the function is increasing or decreasing, we set the first derivative equal to zero and solve for $x$ : $0 = -4x^3 + 12x^2$ $0 = -4x^2(x - 3)$ So, $x = 0$ and $x = 3$ are our critical points.

STEP 7

STEP 8

Since the function goes from increasing to decreasing at $x = 3$ , we have a relative maximum there.
Let's find the y-value: $y = -(3)^4 + 4(3)^3 = -81 + 108 = 27$ .
So, the relative maximum is at $(3, 27)$ .

STEP 9

We need the second derivative to talk about concavity: $y'' = -12x^2 + 24x$

STEP 10

We set the second derivative equal to zero and solve for $x$ : $0 = -12x^2 + 24x$ $0 = -12x(x - 2)$ So, $x = 0$ and $x = 2$ are our potential inflection points.

STEP 11

Let's test $x = -1$ , $x = 1$ , and $x = 3$ : * For $x = -1$ , $y'' = -12(-1)^2 + 24(-1) = -12 - 24 = -36 < 0$ , so concave down. * For $x = 1$ , $y'' = -12(1)^2 + 24(1) = -12 + 24 = 12 > 0$ , so concave up. * For $x = 3$ , $y'' = -12(3)^2 + 24(3) = -108 + 72 = -36 < 0$ , so concave down.

STEP 12

Since the concavity changes at both $x = 0$ and $x = 2$ , these are inflection points.
We already know the y-value for $x = 0$ is $0$ .
For $x = 2$ , $y = -(2)^4 + 4(2)^3 = -16 + 32 = 16$ .
So, our inflection points are $(0, 0)$ and $(2, 16)$ .

STEP 13

STEP 14

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Studdy Solution

STEP 1

STEP 2

1. Find the intercepts2. Analyze increasing/decreasing behavior and extrema3. Analyze concavity and inflection points4. Sketch the graph

STEP 3

To find the y-intercept, we **set** x=0x = 0x=0 in our function: y=−(0)4+4(0)3=0y = -(0)^4 + 4(0)^3 = 0y=−(0)4+4(0)3=0.So, the y-intercept is at the point (0,0)(0, 0)(0,0).Boom!

STEP 4

STEP 5

First, we need the **first derivative** of our function, which tells us about the slope: y′=−4x3+12x2y' = -4x^3 + 12x^2y′=−4x3+12x2

STEP 6

To find where the function is increasing or decreasing, we **set** the first derivative equal to zero and **solve** for xxx: 0=−4x3+12x20 = -4x^3 + 12x^20=−4x3+12x2 0=−4x2(x−3)0 = -4x^2(x - 3)0=−4x2(x−3)So, x=0x = 0x=0 and x=3x = 3x=3 are our **critical points**.

STEP 7

STEP 8

Since the function goes from increasing to decreasing at x=3x = 3x=3, we have a **relative maximum** there.Let's find the y-value: y=−(3)4+4(3)3=−81+108=27y = -(3)^4 + 4(3)^3 = -81 + 108 = 27y=−(3)4+4(3)3=−81+108=27.So, the relative maximum is at (3,27)(3, 27)(3,27).

STEP 9

We need the **second derivative** to talk about concavity: y′′=−12x2+24xy'' = -12x^2 + 24xy′′=−12x2+24x

STEP 10

We **set** the second derivative equal to zero and **solve** for xxx: 0=−12x2+24x0 = -12x^2 + 24x0=−12x2+24x 0=−12x(x−2)0 = -12x(x - 2)0=−12x(x−2)So, x=0x = 0x=0 and x=2x = 2x=2 are our **potential inflection points**.

STEP 11

STEP 12

STEP 13

STEP 14

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1. Find the intercepts
2. Analyze increasing/decreasing behavior and extrema
3. Analyze concavity and inflection points
4. Sketch the graph

To find the y-intercept, we set $x = 0$ in our function: $y = -(0)^4 + 4(0)^3 = 0$ .
So, the y-intercept is at the point $(0, 0)$ .
Boom!

First, we need the first derivative of our function, which tells us about the slope: $y' = -4x^3 + 12x^2$

To find where the function is increasing or decreasing, we set the first derivative equal to zero and solve for $x$ : $0 = -4x^3 + 12x^2$ $0 = -4x^2(x - 3)$ So, $x = 0$ and $x = 3$ are our critical points.

Since the function goes from increasing to decreasing at $x = 3$ , we have a relative maximum there.
Let's find the y-value: $y = -(3)^4 + 4(3)^3 = -81 + 108 = 27$ .
So, the relative maximum is at $(3, 27)$ .

We need the second derivative to talk about concavity: $y'' = -12x^2 + 24x$

We set the second derivative equal to zero and solve for $x$ : $0 = -12x^2 + 24x$ $0 = -12x(x - 2)$ So, $x = 0$ and $x = 2$ are our potential inflection points.