Question

Please help!!
Consider the function f(x)=(3x^2)-(2x^-1)-(3x^1/3)+2
Determine the equation of the line tangent to y=f(x) at x=-1

Answer 1

Do you know how to take the derivative?

Answer 2

Yes!

Answer 3

Show me!

Answer 4

I used the power rule and got - f'(x)= (6x)+(2x^-2)-(x^-2/3)

Answer 5

you should get \[y'=2/x^2-1/x^{2/3}+6 x\]

Answer 6

yep same thing!

Answer 7

Okay, so then how do I do the other question? :S

Answer 8

So with the derivative we found, we can find the SLOPE of the line at x = -1

Answer 9

f'(-1)= (6(-1))+(2(-1)^-2)-((-1)^-2/3) =???

Answer 10

tell me what you get for our slope

Answer 11

-5?

Answer 12

Make sure you type it into your calculator properly.. I got -9

Answer 13

I tried a couple time and also asked my friend to try and we both get -5.

Answer 14

hahah oops. my bad, it is -5

Answer 15

from here, we know that the slope of the tangent line is -5, therefore
y=mx+b
y = (-5)x+b
y = (-5)(-1)+b

Notice we still have two unknowns,is there a way we can get a y value from some where?

Answer 16

if we use f(x)=(3x^2)-(2x^-1)-(3x^1/3)+2, where x = -1, we will get an output of the y co ordinate.. so, find f(-1) now

Answer 17

Well I also had to find f(-1) which equals 10. So is that my y?

Answer 18

yes it sure it, it follows that,
y = mx+b
10 = (-5)(-1) + b

Solve for b

Answer 19

Okay, thank you! :)

Answer 20

so your tangent line is y = -5x + 5
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