Question

x^4-3x^2 +2

Write the equation of the line tangent to the graph at x=1

Answer 1

Also to put it out there I didnt learn derivatives sooo i cant use derivatives

Answer 2

\[y=\frac{dy}{dx}_{(1,0)}x+c\]

Answer 3

just when i mentioned it @jiteshmeghwal9

Answer 4

how would u do this without derivatives and etc

Answer 5

\[y=x_1^3x-3x_1x+2\]x_1=1

Answer 6

nope never learned that either

Answer 7

maybe @FaiqRaees can help u

Answer 8

i learned about the diffferencce quotient formula

Answer 9

i learned it this way

Answer 10

What h and x represents in the answer

Answer 11

idk *sigh*

Answer 12

@FaiqRaees so basically no help i suppose

Answer 13

@jay

Answer 14

a slight change @khantahmins... it will be lim h tends to 0 and then the expression you used..

Answer 15

It represents the slope basically

Answer 16

first let y=x^4-3x^2 +2

Answer 17

then the slope at that point will be 4x^3-6x at x=1 i.e. -2

Answer 18

the value of y at x=1 is 0

Answer 19

so the required tangent line is y-0=-2(x-1) i.e. 2x+y=2

Answer 20

@khantahmina

Answer 21

oh sorry @jango_IN_DTOWN i was a bit busy

Answer 22

i didnt get what u meant by "then the slope at that point will be 4x^3-6x at x=1 i.e. -2"

Answer 23

@zepdrix

Answer 24

Hmm, I'm not really sure what else we can do besides taking the limit of the difference quotient. We have a fourth degree term, so it will be a tad bit of work expanding out the stuff. Shouldn't be too bad though.

Answer 25

\[\rm \frac{d}{dx}(x^4-3x^2+2)=\lim_{h\to0}\frac{(x+h)^4-3(x+h)^2+2-(x^4-3x^2+2)}{h}\]

Answer 26

Understand the first step, setting up the difference quotient?

Answer 27

yes
i only understand that part

Answer 28

@zepdrix

Answer 29

so i expanded it all and this is what i got

Answer 30

\[[x^4+4x^3h+6x^@h^2+4xh^3+h^4-3(x^2+2xh+h^2)-2]-(x^4+3x^2+2)/h\]

Answer 31

Mmm ya looks good.
Distributing the -3 and the -1 respectively gives us this,\[\rm \lim_{h\to0}\frac{\color{orangered}{x^4+4x^3h+6x^2h^2+4xh^3+h^4-3x^2-6xh-6h^2-2}-x^4-3x^2-2}{h}\]

Answer 32

So what's going to happen is,
everything that doesn't contain an h should cancel out.

Answer 33

ok lemme work that out n ill tell u what i got

Answer 34

@zepdrix i think u got a mistake

Answer 35

it s supposed to be -3h^2

Answer 36

not -6h^2

Answer 37

\[\rm \lim_{h\to0}\frac{\color{orangered}{x^4+4x^3h+6x^2h^2+4xh^3+h^4-3x^2-6xh-3h^2-2}-x^4-3x^2-2}{h}\]Ah good catch hehe

Answer 38

:)

Answer 39

x^4 cancels out
2 cancels out

Answer 40

Oh we have a problem with the squared term.
We forgot the negative in our original setup I think.\[[x^4+4x^3h+6x^@h^2+4xh^3+h^4-3(x^2+2xh+h^2)-2]-(x^4\color{red}{+}3x^2+2)/h\]

Answer 41

thats it

Answer 42

\[\rm \lim_{h\to0}\frac{\color{orangered}{x^4+4x^3h+6x^2h^2+4xh^3+h^4-3x^2-6xh-3h^2-2}-x^4+3x^2-2}{h}\]So distributing the negative would give us +3x^2 like this.

Answer 43

noo the original had a -3x^2

Answer 44

Yes, that's why I wrote it in red.
Red is bad.

Answer 45

so distributing it would be +3x^2

Answer 46

got it

Answer 47

So the squared terms cancel? Yay!

Answer 48

so \[4x63h+6x^2h+4xh^3+h^4-6xh-3h^2/h\]

Answer 49

4x^3h imeant ^^^

Answer 50

Oh the carot is on the 6 key haha, I see what happened there ;o

Answer 51

\[\large\rm \lim_{h\to0}\frac{4x^3h+6x^2h^2+4xh^3+h^4-6xh-3h^2}{h}\]Ok good.
Next, divide an h out of everything, ya?

Answer 52

ok

Answer 53

Squared h on your second term*

Answer 54

where

Answer 55

6x^2h^2?

Answer 56

ya

Answer 57

so i have \[h(4x^3+6x^2+4xh^2+h^3-6x-3h) /h\]

Answer 58

ok imm a little confused here n impatient how is this going to help me find the equation of th e line tangent to x=1

Answer 59

No no you forgot the h^2 on the second term again :O

Answer 60

You should still have an h on that second term after factoring.\[h(4x^3+6x^2\color{royalblue}{h}+4xh^2+h^3-6x-3h) /h\]

Answer 61

im soo lost

Answer 62

So let's talk about what the difference quotient is...
It's just your slope formula.

It let's you find the slope between some point x, and some other point which is distance h away from the first point.