Thread: How to do this (tangent line)

I am beginning to study for my Calc 1 final by doing all of the old finals that are up on the math dept. page.

One question is:

If the line tangent to the graph of f (x) = x3 + 1 at the point (a, f (a)) passes
through the origin, then a is:

How am I to go about doing this? In all the practice I have done, and all the questions that have shown on my exam, a was given when it asked for a tangent line passing through a point (a,f(a)). So I don't know how to find a........

a=0 since the origins coordinates are (0,0)

It's a MC question, and none of the answers are 0. That's why I am confused...

What are the choices?

A) $\displaystyle sqrt 3$
B) $\displaystyle 1/sqrt 3$
C)$\displaystyle cbrt 2$
D) $\displaystyle 1/cbrt 2$

I have no clue. I still think it is 0.

You and me both.

What is the answer supposed to be?

Nah, I don't think it's 0. I believe it is D.

$\displaystyle f(x) = x^3 + 1$

$\displaystyle f'(x) = 3x^2$

If the line tangent to $\displaystyle f(x)$ is in the form $\displaystyle y = mx + b$, we know $\displaystyle b = 0$ since it passes through the origin. So all we know is the slope of this line, $\displaystyle m = 3x^2$. So the equation of our tangent line is:

$\displaystyle y = (3x^2)x = 3x^3$. We wish to find when this intersects the original function. So we equate them.

$\displaystyle 3x^3 = x^3 + 1$

$\displaystyle 2x^3 = 1$

$\displaystyle x^3 = \frac{1}{2}$

$\displaystyle x = \frac{1}{2^{1/3}}$

So choice D it is.

Cool stuff. Thanks!

I love how we never learned such a thing, and it was on a final in 08. Oh boy.

Nvm

10char