Question

Let \displaystyle f(x) = 4x\sqrt{2-3x}.
Find the equation of line tangent to the graph of f at x = -2.

not sure how to do this.

Answer 1

Ok so we want an equation for the line tangent to the graph of f.
It will be of the form,
\[\Large\bf\sf y_{\tan}\quad=\quad \color{orangered}{m}x+b\]
We only have to find a couple of things, the `slope` and the `y-intercept`.

Answer 2

The derivative evaluated at x=-2 will give us our slope.
\[\Large\bf\sf \color{orangered}{m\quad=\quad f'(-2)}\]

Answer 3

Do you know how to find the slope of f?

Answer 4

I calculated (or i attempted to calculate) f'(-2) and I got 16(sqrt(2)) but I wasn't sure how to plug that in without b or y since I need y=mx+b.

Answer 5

If that's correct (I'll check it in a sec :D ), then we plug it in for our m.
Then to find b, we need a coordinate pair that we can plug into the tangent line.

The function and tangent line will share the coordinates where the tangent line meets the curve.

So find a coordinate pair.
f(-2) = ?

Answer 6

I think it's wrong lol I think I forgot to make it -2 when I subbed in.

Answer 7

wait no. i put in -2

Answer 8

Hmm I came up with 11sqrt2.
Lemme try that again real quick and make sure I didn't mess up somewhere :U

Answer 9

it's very possible I'm the one who made the mistake I'm not too good with this

Answer 10

\[\Large\bf\sf f(x)\quad=\quad 4x\sqrt{2-3x}\]
\[\Large\bf\sf f'(x)\quad=\quad 4\sqrt{2-3x}+4x\frac{-3}{2\sqrt{2-3x}}\]

Evaluating at -2,\[\Large\bf\sf f'(-2)\quad=\quad 4\sqrt{8}+\frac{24}{2\sqrt{8}}\]That look correct so far? :o

Answer 11

can you tell me how you got the 3 in the numerator? because I had 1/(2sqrt(2-3x))

Answer 12

for square roots isn't it 1/2sqrtx?

Answer 13

This is a good derivative to memorize:\[\Large\bf\sf \frac{d}{dx}\sqrt{x}\quad=\quad \frac{1}{2\sqrt x}\]

Yes very good. Looks like you remembered that one :)
Just have to chain rule after that, multiplying the derivative of the inner function.

Answer 14

what i did was d/dx (4x)(sqrt2-3x)+(4x)d/dx(sqrt2-3x) and so I assumed that the derivative of sqrt of 2-3x was just (1/(2)(sqrt(2-3x)) I'll have to look back at my notes about the chain rule because I must have forgot

Answer 15

\[\Large\bf\sf \frac{d}{dx}\sqrt{2-3x}\quad=\quad \frac{1}{2\sqrt{2-3x}}\cdot\frac{d}{dx}(2-3x)\]Yah we can't forget that chain rule!! :D

Answer 16

so you put the 2-3x in the numerator too?

Answer 17

or. derivative of 2-3x

Answer 18

that's what i meant. derivative.

Answer 19

Yes, the derivative, to save space :)

Answer 20

okay that makes sense, i see where the 3 comes in haha.

Answer 21

Then to find b, you need a coordinate pair.\[\Large\bf\sf f(-2)\quad=\quad 4(-2)\sqrt{2-3(-2)}\quad=\quad ?\]

Answer 22

so would that be \[16\sqrt{2}\] ?

Answer 23

Mmm yes! But with a negative in front right?

Answer 24

yes sorry! i have it in my notes forgot to add that lol

Answer 25

\[\Large\bf\sf y_{\tan}\quad=\quad 11\sqrt2 x+b\]So to find the y-intercept you plug in the coordinate we found.

( -2, -16sqrt2 )

Answer 26

so i would put \[16\sqrt{2}\] for b?

Answer 27

-16*

Answer 28

Hmm I think it's \(\Large\bf\sf b=6\sqrt2\)
Might wanna check your work again :O

Answer 29

i was just asking what to plug in. So since ( -2, -16sqrt2 ) are the coordinates, would i plug in -2 for x and then 16sqrt2 for y and solve for b?

Answer 30

Oh yes +_+

Answer 31

okay i see how you got b now and i also got \[6\sqrt{2}\]

Answer 32

so the equation would be \[y=11\sqrt{2} +6\sqrt{2}\]?

Answer 33

With an x in there somewhere, yes? :)

Answer 34

yes yes i keep forgetting stuff haha. so \[y=11\sqrt{2}x+6\sqrt{2}\]

Answer 35

Mmmm yah looks right
Yay! Good job! \c:/

Answer 36

thank you so much for your help!

Answer 37

no prob! :o
Oh also I noticed you've only been here a lil while,

\(\Large\bf \color{#8C3D62 }{\text{Welcome to OpenStudy! :)}}\)

Answer 38

thanks :) It's one of the best things I've discovered on the internet!