Question

find all tangent lines to y = cubed root of x that have a slope of 3/4

Answer 1

so you know you need to solve the equation y'=3/4 for x?

Answer 2

derivative you should think slope of tangent line
that is why you are solving y'=3/4

Answer 3

I'm not getting it.

Answer 4

you are given the slope of the tangent line is 3/4
you know to find the slope of the tangent line you must find the derivative?

Answer 5

Yes

Answer 6

so what is the problem?

Answer 7

differentiating y ? or what is it that you aren't getting in?

Answer 8

you can write your y as x^(1/3) and use power rule

Answer 9

\[(x^n)'=n x^{n-1}\]

Answer 10

\[(x^\frac{1}{3})'=?\]

Answer 11

(x/3)^-2/3

Answer 12

the 1/3 part shouldn't be raised to the (-2/3)

Answer 13

Yeah, I didn't intend that.

Answer 14

\[(x^n)'=nx^{n-1} \\ (x^\frac{1}{3})'=\frac{1}{3}x^{\frac{1}{3}-1}=\frac{1}{3}x^{\frac{-2}{3}}\]

Answer 15

\[y=x^\frac{1}{3} \\ y'=\frac{1}{3 x^{\frac{2}{3}}} \\ \text{ so remember we needed to solve } y'=\frac{3}{4} \text{ for } x \\ \text{ so we need to solve } \frac{1}{3x^\frac{2}{3}}=\frac{3}{4} \text{ for } x \]

Answer 16

Why?

Answer 17

i don't understand the y'=3/4

Answer 18

OH

Answer 19

It just clicked

Answer 20

to find for which x values on the graph of f(x)=x^(1/3) we have the slopes of their tangent lines is 3/4

Answer 21

I understand now

Answer 22

to find the tangent line we just need the slope and the point
we already have the slope we just the need the point
and remember the tangent line will look like this:
\[y-f(a)=f'(a)(x-a) \\ \text{ where we already know } f'(a) \text{ to be } \frac{3}{4} \\ y-f(a)=\frac{3}{4}(x-a)\]

Answer 23

so that is what your answer will look like
and you should have two answers

Answer 24

Ok, how do I find X

Answer 25

you still having trouble solving that equation for x
one sec let me repost it again

Answer 26

\[y=x^\frac{1}{3} \\ y'=\frac{1}{3 x^{\frac{2}{3}}} \\ \text{ so remember we needed \to solve } y'=\frac{3}{4} \text{ for } x \\ \text{ so we need \to solve } \frac{1}{3x^\frac{2}{3}}=\frac{3}{4} \text{ for } x \\ 3x^\frac{2}{3}=\frac{4}{3} \text{ just flipped both sides } \\ x^\frac{2}{3}=\frac{4}{9} \text{ just divided both sides by } 3 \\ \text{ you have even power so you will have two answers here } \\ x= \pm (\frac{4}{9})^\frac{3}{2}\]
this answer can be simplified a bit

Answer 27

I suggest doing the square root part before doing the cubing

Answer 28

So what is y1

Answer 29

y1?
are you talking about y' ?

Answer 30

or are you talking about the corresponding y values for the x values for which the tangent lines with slope 3/4 occurs?

Answer 31

In point slop formula the y1
as in y - y1 = m(x-x1)

Answer 32

ok so have are having trouble pluggin into y=x^(1/3) ?

Answer 33

you should have gotten the x values are -8/27 and 8/27

Answer 34

\[y=(\pm \frac{8}{27})^\frac{1}{3}=\pm (\frac{8}{27})^\frac{1}{3}\]
just take the cube root of the numerator and denominator

Answer 35

\[y=\pm \frac{2}{3}\]

Answer 36

so tangent lines with slope 3/4 will go through the following two points on y=f(x):
(-8/27,-2/3)
(8/27,2/3)

Answer 37

and I don't mean the same tangent line will go through those two points
I'm talking about two separate points on the curve which means we are talking about two separate tangent lines that satisfies the given conditions

Answer 38

ok
so y - 2/3 = 3/4(x- (-8/27))
y + 2/3 = 3/4(x+ (-8/27))

Answer 39

well m=3/4 with as you called it (x1,y1)=(-8/27,-2/3)
means you have y+2/3=3/4(x+8/27) so this line is correct
but m=3/4 with (x1,y1)=(8/27,2/3) gives
y-2/3=3/4(x-8/27)

Answer 40

it was probably just a type-o you made and that is what you meant though

Answer 41

Yep. Thanks again!

Answer 42

i didn't actually see the type-o in your second equation until just now... For some reason I only noticed the first type-o in the first equation.

Answer 43

anyways goodnight