Tangent Line

**Aryth** · January 14th 2012, 01:04 PM

Find the equation of the line tangent to the graph of $f(x) = -4x^3 + 3\sqrt[5]{x} +4$ at the point $(-1,f(-1))$ . Leave your answer in the form $y = mx + b$ .

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 $y = f(a) + f'(a)(x-a)$

Which gave me:

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

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

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.

**Plato** · January 14th 2012, 01:17 PM

In the real numbers $\sqrt[5]{-1}=-1$ .

Thread: Tangent Line

LinkBack

Thread Tools

Display

Tangent Line

Re: Tangent Line

Similar Math Help Forum Discussions

Find the parametric equations of the tangent line to a line (Answer included)

Finding equation of a tangent line given slope of tangent

At what point on the given curve is the tangent line parallel to the line 3x - y = 3?

tangent line & normal line question

[SOLVED] Tangent Line/Secant Line To A Parabola

Search Tags