Question

Math help plz

Answer 1

Find the equation of the tangent line to the graph of \[F(x) = \frac{ 2x }{ x^2+3 }\] at \[(-1, \frac{ -1 }{ 2 })\]. Determine the point(s) on the graph where the tangent line is horizontal.

Answer 2

So i’ve Been given the function and two points. I need to first find the slope of the tangent. Once i have the slope and the points. I can use the equation \[y - y_1 = m( x - x_1)\]

Answer 3

Nvm i figured it out lol

Answer 4

I’ve found the equation of the tangent line. But i can’t do the next question by myself. @Narad

The question is, Determine the point(s) on the graph where the tangent line is horizontal.

Answer 5

Let me post my solution to the first part. One second.

Answer 6

Do i take the second derivative? Or maybe third?

Answer 7

@Nnesha

Answer 8

I would have done the entire write up on Mathcha

Answer 9

Probably wouldn't have used as many words though

Answer 10

Your tangent line is a bit off btw

Answer 11

Omg i see it... i wrote 1/2 instead of 1/4 lol

Answer 12

Correct

Answer 13

When applying the point slope formula
\(y - y_1 = m(x - x_1)\)

you have to distribute the \(m\) after inserting the points
\(y - y_1 = mx - mx_1\)

Answer 14

I see :)

Answer 15

Very good.

Answer 16

So how do you determine the point(s) on the graph where the tangent line is horizontal?

Answer 17

Sounds a second derivative level task.

Answer 18

I think you're supposed to find the second derivative then set it to 0 and solve for x, but don't quote me on it. You're just finding the values for which the slope is zero.

Answer 19

Yes

Answer 20

Ok let me try.

Answer 21

the answer is...

Answer 22

Explain how you got the \(\sqrt{3}\)

Answer 23

that’s the book answer. when i plugged in 0 in to the second derivative *as you told me* i got 1/3.

Answer 24

my posted my solution up there

Answer 25

That 1/3 is your f(x) value. You're supposed to use in place of f(x) for the original function then solve for x.

Answer 26

Since f(x) is an odd function you're supposed to also try the the negative of the x you found to see if that is also produces a slope of zero.

Answer 27

ok i’ll do that... let me finish this implicit notes first. i’ll post the answer here

Answer 28

You're supposed to test them using f'(x)

Answer 29

f'(x) = 0

Answer 30

or rather test if f prime of (1/3) = 0

Answer 31

When you set f(x) = 1/3 you're looking for the x coordinate of the point that the horizontal tangent line intercepts

Answer 32

i see now.

Answer 33

By the way you know that f(x) is odd because you can rewrite it as
\(f(x) = \left(\dfrac{1}{x^2 + 3}\right)(2x)\)

Answer 34

The first expression is even, the second expression is odd. Even times Odd = Odd

Answer 35

Is that the horizontal tangent line the second question is talking about?

Answer 36

yes... the question asks, Determine the point(s) where the tangent line is horizontal

Answer 37

we found the tangent line using the first derivative remember?

Answer 38

Yes, you did. But the question question is talking about a different tangent line that is horizontal.

Answer 39

tbh i don’t get the question.. i don’t see no horizontal line anywhere on the graph

Answer 40

thats why lol

Answer 41

now tell me this. What happens if you graph y = 1/3 and y = -1/3

Answer 42

im not getting what you are trying to say

Answer 43

Sorry, I meant 1/sqrt(3)

Answer 44

y = 1/sqrt(3)

Answer 45

I think you made a mistake in your calculation

Answer 46

That's the horizontal tangent line

Answer 47

Let's research how to find the correct values algebraically if you haven't already

Answer 48

so i was suppose to get 1/sqrt(3), when i plugged in f”(0)??

Answer 49

i have no clue. plz search with me lol

Answer 50

Okay, I got it. That's my bad. We're supposed to set f prime x equal to zero and solve for x

Answer 51

To find the y-values

Answer 52

2nd derivative is for critical points. It's been a while for me

Answer 53

So set the first derivative equal to zero then solve for x

Answer 54

let me confrim once again this time... we’re suppose to find the first derivative equal to zero and solve for x right? lol

Answer 55

ok, i’ll post the answer in a bit.. :)

Answer 56

Yes

Answer 57

i’ve actually already tried that before but erased it lol

Answer 58

That's not how you do it. You set the whole f(x) equal to zero. not just the x

Answer 59

omg i see now /.-

Answer 60

plz work with me on this one

Answer 61

so far so good?

Answer 62

You shouldn't have done the multiplication

Answer 63

yeah idk what am i doing :/

Answer 64

You're supposed to multiply both sides of the equation by the denominator to get rid of it

Answer 65

Then continue solving for x from there

Answer 66

Do you get it? Nope
Start from here
\(\dfrac{-2x^2+6}{(x^2 + 3)^2}=0\)

Answer 67

You get ZERO on the other side of the equal sign because you set \(f(x)\) to ZERO remember?

Answer 68

Anything times zero is zero

Answer 69

So you should be left with \(-2x^2 + 6 = 0\)

Answer 70

omg..

Answer 71

And you continue solving for \(x\) from there

Answer 72

i dont get it??? then how do i take the (x^2+3)^2 to the numerator ??

Answer 73

omg.... /.-

Answer 74

If you have \(\dfrac{x}{3} = 0\) How do you solve for \(x\) in this case? This is basic algebra

Answer 75

i got it...

Answer 76

yup... 0/3 which is just zero....i wasn’t expecting the whole denominator to cancel out like that..

Answer 77

so how do we find the other .....aka negative x-value?

Answer 78

You did not do that correctly. Here are the correct steps:

Answer 79

uhgg /.- what did i do wrong now? lol

Answer 80

oh u mean the + and - sign??

Answer 81

AAh i always forget to put that in...

Answer 82

i mean to say their corresponding y values

Answer 83

Now you input that in to f( _)

Answer 84

and find the y

Answer 85

\(f(\sqrt{3})\)

Answer 86

wait, let me confirm this... do i plugin that into the first derivative or the original function?

Answer 87

Original function.

Answer 88

AAAh. thanks mate. brb :)

Answer 89

Translate that comment you made in the chat to English

Answer 90

here?

Answer 91

Wherever, as long as I can see the translation

Answer 92

k she asked me *aren’t you studying math a bit too much?* and i said *that poor dude hero’s stil helping me with a math problem.* and then i said *i love him*. that’s it.

Answer 93

Ah okay, but next time, just say it in English.

Answer 94

but english is overrated tho

Answer 95

It is, but it's the only language I know

Answer 96

icy.. i got chew bro.

Answer 97

why am i not getting the right answer?!?!??!?!?!!?!

Answer 98

I would like to see what you've done. You should get the right answer. Show your work like you've always done

Answer 99

Hurry up because I'm going in a minute

Answer 100

yes sir. let me draw my frustrating

Answer 101

Okay, so is that simplified completely?

Answer 102

is this it?

Answer 103

Multiply top and bottom by \(\sqrt{3}\) then simplify again

Answer 104

You multiplied top and botttom by square root three, but for got to "simplify again"

Answer 105

Cancel what is common to top and bottom

Answer 106

isnt sqrt(3)* sqrt(3) = 3?

Answer 107

It is but you have a three in the numerator AND denominator. Do you know what to do with that?

Answer 108

omg that... ofc

Answer 109

Finally

Answer 110

this is the type of tedious math i dont like

Answer 111

and the other one plz one sec

Answer 112

It's only tedious when you're first starting off.

Answer 113

It gets easier the more you do it if you don't forget certain concepts

Answer 114

You get the same value. Except its negative. You don't really need my help with the other one.

Answer 115

You got it. All you have to do is just graph the points to verify and you're done.

Answer 116

You confused yourself

Answer 117

The negative is just -1 it doesn't suddenly get moved to a negative exponent

Answer 118

ooooooo

Answer 119

smh

Answer 120

You keep expanding when you need to be simplifying

Answer 121

There was no need to expand

Answer 122

Rule of exponents
\((ab)^2 = a^2b^2\)

In this case \(a = -1\) and \(b = \sqrt{3}\)

Answer 123

but -1^2 is just 1 tho

Answer 124

But don't confuse yourself
\((-1)^2 = 1\) Only when the negative is in parentheses does it become positive

Answer 125

Which in this case it is in parentheses but there will be cases when the negative is not within paraentheses

Answer 126

are we done or are we done ?

Answer 127

thank you soooooooooooooooo much hero! you’re trully a living hero for me. :)

Answer 128

Yep it is done when you graph the points to verify them.