# Thread: Simple functions - zero of f(x)

1. ## Simple functions - zero of f(x)

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.

2. 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 $f''(0)>0$

3. 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

4. Oh, er, my bad then :S