# Tangent Line

• Jan 14th 2012, 01:04 PM
Aryth
Tangent Line
Find the equation of the line tangent to the graph of $\displaystyle f(x) = -4x^3 + 3\sqrt[5]{x} +4$ at the point $\displaystyle (-1,f(-1))$. Leave your answer in the form $\displaystyle y = mx + b$.

I just want to verify if what I'm doing is correct... It has been awhile since I have done one of these.

I assumed I should use the equation $\displaystyle y = f(a) + f'(a)(x-a)$

Which gave me:

$\displaystyle f(-1) = 8 + 3\sqrt[5]{-1}$
$\displaystyle f'(-1) = 12 + \frac{3}{5}\sqrt[5]{(-1)^4} = \frac{63}{5}$

$\displaystyle y = 8 + 3\sqrt[5]{-1} + \frac{63}{5}(x+1)$
$\displaystyle y = \frac{63}{5}x + (8 + \frac{63}{5} + 3\sqrt[5]{-1})$
$\displaystyle y = \frac{63}{5}x + (\frac{103}{5} + 3\sqrt[5]{-1})$

Here is where I have trouble... I am unsure of the fifth root of -1. I have seen that it could be complex, or it could be 1... And I just can't verify it myself. I'm sure it is easier than I am making it out to be, but any help would be appreciated.
• Jan 14th 2012, 01:17 PM
Plato
Re: Tangent Line
In the real numbers $\displaystyle \sqrt[5]{-1}=-1$.