Symmetry about a vertical line (like x = c, c not zero).

Hi all ----

I actually get part (ii) of this question - but I'm just curious - how can I formally prove the symmetry? The question doesn't ask this but I'm just curious. The green box is the solution.

http://i54.tinypic.com/24maptx.jpg

__Part (ii) ---__ I can see that $\displaystyle \int_0^1 2^{-(x - c)^2}$ reaches maximum when $\displaystyle c = 1/2$.

But it also looks like $\displaystyle 2^{-\left(x - \frac{1}{2})^2\right}$ is symmetric about $\displaystyle x = 1/2$. How would I prove this?

I know how to do this for $\displaystyle f(x) = 2^{-x^2}$ ---

$\displaystyle f(x)$ is symmetric about $\displaystyle x = 0$ because $\displaystyle f(-x) = 2^{-(-x)^2} = f(x) = 2^{-x^2} $ so $\displaystyle f(-x) = f(x)$.

Thanks a lot ---

Re: Symmetry about a vertical line (like x = c, c not zero).

$\displaystyle y = 2^{-\left(x - \frac{1}{2})^2\right}$ is symmetric about x=1/2, but not x=1.

try to translate $\displaystyle y = 2^{-\left(x - \frac{1}{2})^2\right}$ 1/2 unit left

Re: Symmetry about a vertical line (like x = c, c not zero).

Quote:

Originally Posted by

**piscoau** $\displaystyle y = 2^{-\left(x - \frac{1}{2})^2\right}$ is symmetric about x=1/2, but not x=1.

try to translate $\displaystyle y = 2^{-\left(x - \frac{1}{2})^2\right}$ 1/2 unit left

Hi piscoau ---

Thanks - I fixed it now.

But I know how to show symmetry for $\displaystyle f(x) = 2^{-x^2}$. I did this already in my original post. I want to prove symmetry for $\displaystyle 2^{-\left(x - \frac{1}{2})^2\right}$ about $\displaystyle x = 1/2$.

Re: Symmetry about a vertical line (like x = c, c not zero).

do you know how to translate the graph of a function horizontally? If the question want you to prove f(x) is symmetry about x=c, then you need to translate the graph c unit left/right.

For example, you want to prove function (x-1)^4 is symmetry about x=1. you need to translate the function 1 unit left, the function will become (x-1+1)^4 = x^4. Obviously it's symmetry about x=0, i.e. x=0 is its symmetry axis of x^4. If you translate x^4 back to (x-1)^4, the whole graph will move 1 unit right, i.e. the symmetry axis will move 1 unit right too. Therefore, (x-1)^4 is symmetry about x=1.

If a figure is translated either horizontally or vertically, only its position changed, its shape, size, pointing direction DOESN'T change. So I think you know how to do it know.

Re: Symmetry about a vertical line (like x = c, c not zero).

Quote:

Originally Posted by

**mathminor827** Hi piscoau ---

Thanks - I fixed it now.

But I know how to show symmetry for $\displaystyle f(x) = 2^{-x^2}$. I did this already in my original post. I want to prove symmetry for $\displaystyle 2^{-\left(x - \frac{1}{2})^2\right}$ about $\displaystyle x = 1/2$.

1. Take 2 x-values equidistant from $\displaystyle x = \tfrac12$ . for instance $\displaystyle x = \tfrac12 - d~\text{or}~x=\tfrac12 + d~,~d \in \mathbb{R}$

2. If $\displaystyle f\left(\tfrac12 - d \right) = f\left(\tfrac12 + d \right)$

then the graph of f is symmetric about $\displaystyle x = \tfrac12$