Find the slope of the tangent line to the curve

[yanarains]

Find the slope of the tangent line to the curve

f(x)= 1/(1-ax)^n

at the point where x = 0.8.

With a = 9, and n = 3.
Round the answer to three decimal places.

OK, WELL I REALLY DONT THINK I EVEN KNOW HOW TO SET THIS UP BUT I WILL TRY PLEASE SOMEONE CORRECT MY MISTAKES!!!!

f(x)=[9(0.8)]^-3
f(x)=-3[9(0.8)]^-1-3
f(x)=-3[7.2]^-4
f(x)=4.59393658 ( I then will plug this number into the original equation for X)

f(x)=1/[1-9(4.59393658)]^3
=-49639.21983
I know that is completely wrong but this is how I am working the problem out.
Thank again for ALL the help![/u]

 

[stapel]

I REALLY DONT THINK I EVEN KNOW HOW TO SET THIS UP

I'm sorry to hear that you missed the class session(s) where derivatives and tangent lines were covered. :shock:

f(x)=[9(0.8)]^-3

If you plug everything in, you've got a constant function, and the derivative of a constant is zero. So the above cannot be valid.

Find the slope of the tangent line to the curve f(x)= 1/(1-ax)^n at the point where x = 0.8, With a = 9, and n = 3.

So the function is:

. . . . .f(x) = 1 / (1 - 9x)³ = (1 - 9x)^-3

Apply the Power Rule to differentiate. You should end up with:

. . . . .f'(x) = (27) / (1 - 9x)⁴

This gives you the slope of the curve at any point (x, f(x)). Now is when you plug in 0.8 for x. Simplify to find the value of the slope m at the point.

To find the tangent line, you of course need the actual point's coordinates. So plug in 0.8 for x, and find (x, y) = (0.8, f(0.8)). Once you have the point and the slope, you can find the straight-line equation, just like you learned back in algebra. :wink:

Hope that helps!

Eliz.