Question

find the equation of the tangent line to the graph of f at the given point f(x)=x^3 derivative long way

Answer 1

Long way?
Using the limit definition?

Answer 2

\[\Large\rm \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\]

Do you understand what to plug in for f(x+h) and f(x)?

Answer 3

\(\bf f({\color{brown}{ x}})={\color{brown}{ x}}^3 \qquad f({\color{brown}{ x+h}})=({\color{brown}{ \square }})^3\)

Answer 4

yes, x^3 for x+h to the third and x^3 in f(x) and they are on the points [2,8}

Answer 5

i have it written like this (2+deltax)^3 -(2)^3 divided by x

Answer 6

hmm?

Answer 7

Oh so we're evaluating this at x=2? ok

Answer 8

you're just looking for the derivative using the "definition of a limit" right?

Answer 9

\[\Large\rm \lim_{\Delta x\to 0}\frac{(2+\Delta x)^3-(2)^3}{\Delta x}\]Do you understand how to expand out the cube?

Answer 10

i know that you solve that, then cancel any terms that are the same number but have different signs like - or +, but after that i check to see my answer and it is 12x-16

Answer 11

thats the books answer and i get something much different

Answer 12

Yah that answer looks correct.
Let's see where we went wrong...

Answer 13

So expanding the cube gives us:\[\Large\rm \lim_{\Delta x\to 0}\frac{8+12\Delta x+6\Delta x^2+\Delta x^3-8}{\Delta x}\]Which simplifies,\[\Large\rm \lim_{\Delta x\to 0} 12+6\Delta x+\Delta x^2\]Look ok so far?

Answer 14

yes but why does delta x get removed

Answer 15

The 8's subtract,\[\Large\rm \lim_{\Delta x\to 0}\frac{12\Delta x+6\Delta x^2+\Delta x^3}{\Delta x}\]Then we're dividing a Delta x out of each term,\[\Large\rm \lim_{\Delta x\to 0}\frac{12\cancel{\Delta x}+6\Delta x^{\cancel{2}1}+\Delta x^{\cancel{3}2}}{\cancel{\Delta x}}\]

Answer 16

oh ok, you just cancelled the delta x's first ok

Answer 17

The idea with limits is,
We're trying to get them to a point where we can plug in the limiting value without having any problems.

Before, when we had our Delta x in the denominator, plugging in zero caused a problem.
Now that we've simplified it, we can plug our zero directly in!\[\Large\rm \lim_{\Delta x\to 0} 12+6\Delta x+\Delta x^2=12+0+0\]Understand how that works?

Answer 18

yes, i see

Answer 19

but then how does the x come together with the 12

Answer 20

So we have some function \(\Large\rm f(x)\).
The derivative of that function at x=2 represents the `slope` of the line tangent to f(x) there.

So we're looking for some linear function which is tangent to f(x) at x=2.
\[\Large\rm y=\color{orangered}{m}x+b\]And we said that the `slope` of our tangent line is given by the derivative function.\[\Large\rm y=\color{orangered}{f'(2)}x+b\]

Answer 21

And we did all that hard work to find out that f'(2)=12.
That was the long process.\[\Large\rm y=\color{orangered}{12}x+b\]

Answer 22

Since this tangent line `touches` the function at x=2,
It means that the function and our line share the coordinate point there.

So they both pass through (2,8).
That means we can plug this point into our tangent line to solve for b, the y-intercept.

Answer 23

thank you very much very helpful