# Functions and the IVT

Let $g, h$ be continuous functions defined on some interval $J$ and suppose that $g(x) \neq 0$ for any $x \in J$. If $g(x)^2 = h(x)^2$ for all $x \in J$, show that either $g(x) = h(x)$ for all $x \in J$ or $g(x) = -h(x)$ for all $x \in J$.

I know that when I get a square root of something, then the result could be negative of positive. SO I was thinking in doing cases, but I am not really sure where should I add the Intermediate Value Theorem.

Also, $g(x)^2 - h(x)^2 = (g(x) + h(x))(g(x) - h(x))$. Could I make each of them $=0$, and I am done?

Do you need the intermediate value theorem? Let $g(x)$ and $h(x)$ be continuous functions defined on the interval $I$ such that $g(x)^2 = h(x)^2$ and $g(x) \neq 0$ for all $x \in I$. Since $g(x)^2 = h(x)^2$, we know that $1 = (h(x)/g(x))^2$. Taking square roots, we conclude that either $1 = h(x)/g(x)$ or $-1 = h(x)/g(x)$. This is equivalent to the statement that either $g(x) = h(x)$ or $g(x) = -h(x)$.

• exactly what I was thinking!
– user99638
Nov 16, 2013 at 23:59

Hint: Since $g(x)\ne 0$ for all $x$, we can consider the function $\frac{h(x)}{g(x)}$.

(The condition $g(x)\ne 0$ is indeed necessary as $J=[-1,1]$, $g(x)=|x|$, $h(x)=x$ shows)

• How could I apply the IVT here?
– user99638
Nov 16, 2013 at 23:25
• Once I have $(g(x)+h(x))(g(x)-h(x))=0$, then by the IVT, there is some $g(a)=0$, such that $(g(x)-h(x))=0$ or $g(x)+h(x)=0$.
– user99638
Nov 16, 2013 at 23:35