Question

Determine the equation of the tangent to the graph of the function \(f(x)=x^3\) at the point \(\large \frac{1}{2}, \frac{1}{8}\)

Answer 1

calculus?

Answer 2

Yeah

Answer 3

let's suppose tangent is y=mx+b

Answer 4

do u know how to get m?

Answer 5

The tanget of a function is given by its derivate. The derivate of each point is the tangent of the original function on that given point.

So taking the first derivative:
\[f'(x) = 3x^2\]
At the point x = 1/2 its value is
\[f'(1/2) = { 3 \over 4}\]

Answer 6

So I differentiate \(\large y=x^3\)?

Answer 7

yes just like what soati did

Answer 8

according to the soati's statement m=3/4

Answer 9

Why did he substitute \(\frac{1}{2}\) into f'(x)?

Answer 10

then you already have x and y for the point you want, just plug them in to find b

Answer 11

because we want to find tangent at this point

Answer 12

Isn't f'(x) represent the slope of all points on the graph?

Answer 13

Ohh

Answer 14

slope of the tangent is the derivative of the function at that point

Answer 15

now what can we do for get b (y=3/4x+b)

Answer 16

Is the answer \(\large y=\frac{3}{4}x-\frac{1}{4}\)?

Answer 17

yup

Answer 18

Thank you guys for clarifying this! :)

Answer 19

Also, this tangent crosses the main function at another place.

Answer 20

I need to find where; So I set up an equality;
\(\large x^3=\frac{3}{4}x-\frac{1}{4}\)

Answer 21

I got \(x^3-3x+1=0\)

Answer 22

How do I factor that out?

Answer 23

use ruffini. try dividing by x +1

Unless I got it wrong, the result should be 4x^2 - 4x + 1
So you will have (x+1)(4x^2 - 4x + 1) = 0

Answer 24

\[x^3-\frac{3}{4}x+\frac{1}{4}=0 \\ x^3-\frac{1}{4}x-\frac{1}{2}x+\frac{1}{4}=0 \\ x(x^2-\frac{1}{4})-\frac{1}{2}(x-\frac{1}{2})=0 \\ x(x-\frac{1}{2})(x+\frac{1}{2})-\frac{1}{2}(x-\frac{1}{2})=0 \\ (x-\frac{1}{2})(x^2+\frac{1}{2}x-\frac{1}{2})=0\]

Answer 25

by the way, your equation should be 4x^3 - 3x + 1 = 0

Answer 26

I see

Answer 27

Ok good! I got it, thank you so much! @mukushla I can't give out 2 medals :$, but thank you both! :)

Answer 28

your welcome