Review question

# What does it mean if a function $f(x)=f(2\alpha-x)$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5090

## Question

In this question we fix a real number $\alpha$ which will be the same throughout. We say that a function $f$ is bilateral if $f(x) = f(2\alpha - x)$ for all $x$.

1. Show that if $f(x) = (x-\alpha)^2$ for all $x$ then the function $f$ is bilateral.

2. On the other hand, if $f(x) = x-\alpha$ for all $x$ then the function $f$ is not bilateral.

3. Show that if $n$ is a non-negative integer and $a$ and $b$ are any real numbers then $\int_a^b x^n \:dx = -\int_b^a x^n \:dx.$

4. Hence show that if $f$ is a polynomial (and $a$ and $b$ are any reals) then $\int_a^b f(x) \:dx = -\int_b^a f(x) \:dx.$

5. Suppose that $f$ is any bilateral function. By considering the area under the curve $y = f(x)$ explain why for any $t \geq \alpha$ we have $\int_\alpha^t f(x) \:dx = \int_{2\alpha-t}^\alpha f(x) \:dx.$

If $f$ is a function we write $G$ for the function defined by $G(t) = \int_\alpha^t f(x) \:dx$ for all $t$.

1. Suppose now that $f$ is any bilateral function. Show that $G(t) = -G(2\alpha-t)$ for all $t$.

2. Suppose that $f$ is a bilateral polynomial such that $G$ is also bilateral. Show that $G(x) = 0$ for all $x$.