How to do this (tangent line)

**Marconis** · Dec 1st 2010, 05:44 PM

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........

**dwsmith** · Dec 1st 2010, 05:50 PM

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

**Marconis** · Dec 1st 2010, 05:52 PM

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

**dwsmith** · Dec 1st 2010, 05:54 PM

What are the choices?

**Marconis** · Dec 1st 2010, 06:02 PM

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

**dwsmith** · Dec 1st 2010, 06:27 PM

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

**Marconis** · Dec 1st 2010, 06:35 PM

You and me both.

**dwsmith** · Dec 1st 2010, 06:35 PM

What is the answer supposed to be?

**elemental** · Dec 1st 2010, 06:36 PM

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

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

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

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

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

$3x^3 = x^3 + 1$

$2x^3 = 1$

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

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

So choice D it is.

**Marconis** · Dec 1st 2010, 06:37 PM

Cool stuff. Thanks!

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

**Marconis** · Dec 1st 2010, 06:39 PM

Nvm

10char

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