Simple functions - zero of f(x)

**lantern4christ** · September 28th 2009, 03:32 AM

I have precalculus in college. I am trying to figure out a definitive formula to answer these equations. I'm not sure if they have a formula or if you kind of just go with it. Help would be appreciated. This is a 2-step problem. (a) find the zero of the function. (b) find the min/max of the parabola.

**Defunkt** · September 28th 2009, 03:57 AM

Originally Posted by lantern4christ

I have precalculus in college. I am trying to figure out a definitive formula to answer these equations. I'm not sure if they have a formula or if you kind of just go with it. Help would be appreciated. This is a 2-step problem. (a) find the zero of the function. (b) find the min/max of the parabola.

To find the zero of a function $f(x)$ , we simply need to solve $f(x) = 0$ with regards to $x$ . That is, find $x_0$ such that $f(x_0) = 0$ .

To find the min/max of a parabola, we solve $f'(x) = 0$ with regards to x. Then, for those values of x we find $f''(x)$ . Say $f'(x_0)=0$ . If $f''(x_0)>0$ then $(x_0,f(x_0))$ is a minimum point. If $f''(x_0)<0$ , then $(x_0,f(x_0))$ is a maximum point.

Example:
Let $f(x) = x^2 - 9$ . Find the zeroes of $f(x)$ and find its min/max points.

First we'll find the zeroes. Solve $x^2-9=0$ :
$x^2-9 = 0 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3$
So $f(3) = f(-3) = 0$ .

Now we'll find the min/max points. First we find $f'(x)$ :
$f'(x) = \frac{d}{dx}(x^2-9) = 2x$

Solve $f'(x) = 0$ : $2x = 0 \Rightarrow x = 0$
So $x=0$ is a critical point. Find $f''(x)$ :
$f''(x) = \frac{d}{dx}(2x) = 2 \Rightarrow f''(0) = 2$

So $(0,f(0)) = (0,-9)$ is a minimum point, since $f''(0)>0$

**CaptainBlack** · September 28th 2009, 04:21 AM

Originally Posted by Defunkt

To find the min/max of a parabola, we solve $f'(x) = 0$ with regards to x.

Not in pre-calculus we don't, here we rely on the properties of the parabola (see this for more information).

CB

**Defunkt** · September 28th 2009, 04:42 AM

Oh, er, my bad then :S

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