graphing

[whiteti]

Sketch the graph of f(x)= x⁴ - 8x² - 9 = (x-3)(x+3)[x² + 1] by first finding its
a) Y-intercept
b)X-intercept(s)
c) derivative
d) critical numbers
e) intervals of increase and decrease
f) local extrema
g) second derivative
h) intervals of concavity
i) inflection points
j) sketch its graph
if it does not have an item, include a statement to that effect

y intercept = -9
x-intercepts, i know you set it to zero, im having trouble factoring, help?
derivative = 4x³ -16x?

the second part of this equation after the second equal sign is throwing me off, what does it mean?

 

[daon2]

x-intercepts, i know you set it to zero, im having trouble factoring, help?
derivative = 4x³ -16x?

the second part of this equation after the second equal sign is throwing me off, what does it mean?

The second equal sign is telling you how f(x) factors, and will help you find your intercepts. For example if f(x)=x^2-2x-3=(x-3)(x+1), the the x-intercepts arre x=+3, x=-1.

Your first derivative is correct. To find the critical numbers, set f'(x)=0 and solve for x.

 

[stinajeana]

The second equal sign is telling you how f(x) factors, and will help you find your intercepts. For example if f(x)=x^2-2x-3=(x-3)(x+1), the the x-intercepts arre x=+3, x=-1.

Your first derivative is correct. To find the critical numbers, set f'(x)=0 and solve for x.

Would critical numbers be 0 and 14?

 

[stinajeana]

Would critical numbers be 0 and 14?

never mind... just realized what I did wrong! I used the quadratic equation when I should have just factored the derivative and made it equal to zero.

 

[stinajeana]

My answer is different.

How did you get those numbers?

I blanked and thought I could use the quadratic formula. So,

4x^3-16x
4x(x^2-4)
4x(x-2)(x+2)

critical points are -2, 0, 2

 

[stinajeana]

My answer is different.

How did you get those numbers?

Now i'm just having trouble finding the intervals of increase and decrease and local extrema. This is what I have so far

 

[Deleted member 4993]

Now i'm just having trouble finding the intervals of increase and decrease and local extrema. This is what I have so far
View attachment 2941

From -∞ < x < -2, does the function increase or decrease?

From -2 < x < 0, does the function increase or decrease?

From 0 < x < ∞, does the function increase or decrease?

What happens to the derivative at local maxima/minima?

 

[whiteti]

From -∞ < x < -2, does the function increase or decrease?

From -2 < x < 0, does the function increase or decrease?

From 0 < x < ∞, does the function increase or decrease?

What happens to the derivative at local maxima/minima?

I'm having trouble finding the second derivative, not really sure how

the first derivative is f'(x)= 4x³ - 16x

 

[srmichael]

I'm having trouble finding the second derivative, not really sure how

the first derivative is f'(x)= 4x³ - 16x

The second derivative is the derivative of the first derivative so.....

 

[whiteti]

The second derivative is the derivative of the first derivative so.....

So i got f''(x) = 4(3x-4) as my final number, how do i get my critical numbers from this?

 

[srmichael]

So i got f''(x) = 4(3x-4) [INCORRECT] as my final number, how do i get my critical numbers from this?

f''(x) = 12x²-16

Setting the second derivative = 0 and where it is undefined (if applicable) gives you potential Points of Inflections. Potential critical numbers come from setting the first derivative = 0 and finding where the first derivative is undefined (if applicable).

 

[JeffM]

So i got f''(x) = 4(3x-4) as my final number, how do i get my critical numbers from this?

You were told way back when to find the critical values by finding the values of x that equate the first derivative to zero.

Your calculation of the second derivative is wrong.

 

[whiteti]

You were told way back when to find the critical values by finding the values of x that equate the first derivative to zero.

Your calculation of the second derivative is wrong.

ok then im lost on how to get it right

 

[Deleted member 4993]

So i got f''(x) = 4(3x-4) as my final number, how do i get my critical numbers from this?

Incorrect:

f'(x) = 4*x³ - 16*x

f"(x) = 4*(3)*x^(3-1) - 16 * x^(1-1) = 12*x² - 16*x⁰ = 12*x² - 16

What do you mean by - how do i get my critical numbers from this?

What is a critical number? Where have you been asked to find that?

 

[Deleted member 4993]

ok then im lost on how to get it right

Read all the answers to the post - Carefully!!

 

[whiteti]

Incorrect:

f'(x) = 4*x³ - 16*x

f"(x) = 4*(3)*x^(3-1) - 16 * x^(1-1) = 12*x² - 16*x⁰ = 12*x² - 16

What do you mean by - how do i get my critical numbers from this?

What is a critical number? Where have you been asked to find that?

Yea i got that answer, but dont i have to factor is or something to get the critical numbers (partition numbers) to creat my sign chart so i know the intervals of increase and decrease and to find the concavity?

 

[HallsofIvy]

Sketch the graph of f(x)= x⁴ - 8x² - 9 = (x-3)(x+3)[x² + 1] by first finding its
a) Y-intercept
b)X-intercept(s)
c) derivative
d) critical numbers
e) intervals of increase and decrease
f) local extrema
g) second derivative
h) intervals of concavity
i) inflection points
j) sketch its graph
if it does not have an item, include a statement to that effect

y intercept = -9

Yes, that is correct.

x-intercepts, i know you set it to zero, im having trouble factoring, help?

???? It's already factored!

derivative = 4x³ -16x?

Yes, that is correct.

the second part of this equation after the second equal sign is throwing me off, what does it mean?

Are you saying that you do not know what "=" means? It means that the two, the original polynomial and the factored form are equal. That is the factored form!