Question

find the equation of the tangent line to the graph f(x)=3x^3-10

Answer 1

any particular point on the curve?

Answer 2

what is the slope of f(x)? Slope is the first derivative, right?

Answer 3

3?

Answer 4

differentiate f(x) = 3x^3 - 10...

f ' (x) = 3(3)x^2 = 9x^2

Answer 5

Ok. I forgot to write "at the point (2,14)" how does that change it?

Answer 6

YES!

You need to know where along f(x) you want to find that tangent line. Any tangent to that f(x) curve will have the slope given by f ' (x), but in order to actually write a line equation, typically you write it for one specific line equation, in other words, tangent at some specific point along f(x) defined by the x-value.

Answer 7

So, at that point (2,14), what is f ' (x) = f ' (2) = ?

Answer 8

2

Answer 9

is that the slope?

Answer 10

No, you need to evaluate that first derivative using x = 2

f ' (x) = 9x^2

f ' (2) = 9(2)^2 = 9(4) = 36

Answer 11

That is the slope of the tangent line to f(x) at the point along f(x) where x = 2

Answer 12

ok. so the eqation is f'(x)=36x?

Answer 13

So now you have a slope of 36 and a point (2,14). You can write a line equation using the point-slope form.

y - y1 = m(x - x1) using point (2,14) as (x1, y1)
and using slope m = f ' (x) = f '(2) = 36

Answer 14

oh! ok! I get it. thank you

Answer 15

no, f ' (x) is the derivative of f(x).... in general, it is just the expression f ' (x) = 9x^2

You can evaluate f ' (x) for any x you want... here, you need to do it at x = 2

Answer 16

For curves, the slope is the first derivative which is a function, not a single number.

Lines have constant slope, so they have a slope m.

Curves have a slope that changes with different x values, so it's an expression using x. But you can solve for slope at any SPECIFIC x by plugging in that x value into the first derivative expression.

Answer 17

ok, i got y=36x-58...

Answer 18

y=36x-58 is correct.

Answer 19

you can check it... is (2,14) a solution to y = 36x - 58?

14 = 36(2) - 58 yes, therefore (2,14) is on the tangent line

Is (2,14) on the curve f(x)? It should be... given in the problem.

Is the slope of the tangent line at x = 2 equal to the derivative of f(x), f ' (x), at f'(x) = 2?

Slope of tangent line is 36 everywhere. Slope of f(x) is f ' (x) = 9x^2, and at f ' (2), it is 36.

So yes :) It all works out! Yay!!

Answer 20

Awesome! Thanks soooooooo much!

Answer 21

Glad to help :) Understand the idea?
1) Find the slope expression using first derivative.
2) Evaluate first derivative at the specific x value to get the slope value (a number).
3) Use point and slope to write equation of tangent line at that particular point.
4) Check work :)

Answer 22

yup. i got it. thanks :D

Answer 23

Great! Works for any question with "find equation of tangent line to some f(x) at point (x1, y1)"

Good luck!

Answer 24

one more quetion. The derivative of theta is zero, right?

Answer 25

uh :) I don't know... is there more to the problem? Theta is often used as a variable for an angle. All alone, it doesn't mean anything specific.

Answer 26

No, i was working on a second problem. it was the same question but for f(theta)=4Sin(theta)-(theta).

Answer 27

oh wait. that would make the answer 1, right?

Answer 28

just for the theta part.

Answer 29

you have to take the derivative of the whole right side... but yes, it looks like it would be "1"

f(theta) = theta is just like f(x) = x, so f '(x) = 1

Just don't forget the 4sin(theta) part ;)

Answer 30

of course :) thanks