Question

determine the point(s) at which the graph of
the function has a horizontal tangent line

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

Answer 1

( 0,0)

Answer 2

how did you solve that?

Answer 3

slope of horizontal line = 0

Answer 4

you can get slope by finding the derivative

Answer 5

so find the derivative of f(x) and see when it equals 0

Answer 6

When you find a derivative of a function (i.e. it's slope at each point), you are finding the tangents (i.e. slopes) at every point. Since the question is asking you to find "horizontal" tangents that equal zero, it's the same as asking you where the slope of this function is zero. Zero slopes for any function occur where minimums or maximums occur, which makes sense, because at these points, the slopes are going from negative to positive, so somewhere in-between there must be a zero slope.

Do you know how to find derivatives?

Answer 7

Yes

Answer 8

Then let's do it, let's find the derivative of this function.

Answer 9

d/dx x^n=nx^n-1

Answer 10

$$
f(x)=\cfrac{x^2}{x^2+1}\\
f'(x)=?
$$
Do you know how to use the quotient rule (bottom, times derivative of the top, minus the top, times the derivative of the bottom, divided by the bottom, squared)?

Answer 11

That formula is correct for a monomial, we need also to apply the quotient rule, using also this monomial derivative.

Answer 12

Let's do this in stops.

What is the denominator multiplied times the derivative of the numerator?

Answer 13

*steps not stops lol

Answer 14

$$
f(x)=\cfrac{x^2}{x^2+1}\\
$$
Step 1 (bottom times derivative of the top):

$$
(x^2+1)\times \cfrac{d}{dx}x^2=?
$$

Answer 15

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

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

f(x) = x^2+1

Answer 16

$$
\cfrac{x^2}{1}+\cfrac{x^2}{x^2}\ne\cfrac{x^2}{x^2+1}\\
$$
This is not equal because,
$$
\cfrac{x^2}{1}+\cfrac{x^2}{x^2}=\cfrac{x^2}{1}\times\cfrac{x^2}{x^2}+\cfrac{x^2}{x^2}\\
=\cfrac{x^4}{1}+\cfrac{x^2}{x^2}=\cfrac{x^4+x^2}{x^2}\ne\cfrac{x^2}{x^2+1}
$$

Answer 17

*correction to second part above:
$$
\cfrac{x^2}{1}+\cfrac{x^2}{x^2}=\cfrac{x^2}{1}\times\cfrac{x^2}{x^2}+\cfrac{x^2}{x^2}\\
=\cfrac{x^4}{\color{red}{x^2}}+\cfrac{x^2}{x^2}=\cfrac{x^4+x^2}{x^2}\ne\cfrac{x^2}{x^2+1}
$$

Answer 18

I'm lost now.

Answer 19

$$
\cfrac{1}{1+2}=\cfrac{1}{3}\ne\cfrac{1}{1}+\cfrac{1}{2}=\cfrac{3}{2}\\
\cfrac{1}{3}\ne\cfrac{3}{2}
$$
This has nothing to do with your problem, but this is what you were starting off with and I want to make sure you understand, but because you need to know this.

Answer 20

Let's now do s step 1:

$$
(x^2+1)\times \cfrac{d}{dx}x^2=?
$$

Answer 21

$$
\cfrac{d}{dx}x^2=2x
$$
So,
$$
(x^2+1)\times \cfrac{d}{dx}x^2=(x^2+1)\times 2x=2x^3+2x
$$
Make sense so far?

Answer 22

yes

Answer 23

That was bottom times derivative of the top.

Step 2 (top times derivative of the bottom):

$$
x^2\cfrac{d}{dx}(x^2+1)=?
$$

Answer 24

$$
\cfrac{d}{dx}(x^2+1)=2x
$$
So,
$$
x^2\cfrac{d}{dx}(x^2+1)=x^2\times 2x=2x^3
$$
Are you okay with this?

Answer 25

Where does the 2x come from

Answer 26

We are doing the quotient rule:

Step 1: bottom times derivative of the top
Step 2: top times derivative of the bottom
Step 3: bottom squared

The we combine the results of each:

$$
\cfrac{\text{Step 1 - Step 2}}{{\text{Step 3}^2}}
$$

Answer 27

The 2x comes from this:
$$
\cfrac{d}{dx}(x^2+1)=2x
$$
This is using your original formula for the derivative of a monomial: d/dx x^n=nx^n-1. Where here, n=2.

Answer 28

$$
\cfrac{d}{dx}(x^2+1)=\cfrac{d}{dx}x^2+\cfrac{d}{dx}1=2\times x^{2-1}+0=2x
$$

Answer 29

questions?

Answer 30

$$
f(x)=\cfrac{x^2}{x^2+1}\\
f'(x)=0\implies\cfrac{(x^2+1)2x-x^2\times2x}{\left (x^2+1\right )^2}=\cfrac{2x}{x^4+2x^2+1}=0\implies x=0
$$
So, horizontal tangents occur at \(x=0\).
Answer: At \(x=0\): \(f(0)=\cfrac{0^2}{0^2+1}=0\).
So, \(f(x)=0\) at the horizontal tangents.