Two functions $f$ and $g$ are said to be equal (written $f=g$) if their domains, codomains and graphs are all equal. A function $f:ℝ\toℝ$ is defined by
\begin{equation*} f(x) = \left\{ \begin{array}{ll} x^2 & \quad x \geq 0 \\ x & \quad x < 0 \end{array} \right. \end{equation*}
a) Sketch the graph of the function $g$ given by $g(x)=f(|x|)$
So the function $g$ is like this?
\begin{equation*} g(x) = \left\{ \begin{array}{ll} (|x|)^2 & \quad x \geq 0 \\ |x| & \quad x < 0 \end{array} \right. \end{equation*}
And the I just graph the right half of the parabola for the first piece and the left half of the V of absolute value function?
b) Let $h:ℝ\toℝ$ be given by $h(x)=x^2$, $x\inℝ$ let $g$ be as defined in part (b). Determine, using the definition of equality whether or not $h=g$. Justify your answer.
In this part I don't know if the part that says "let $g$ be as defined in part (b)" is a mistake and must say part (a) because in part (b) what is defined is $h$ not $g$.
Not sure on how to approach this part to be honest.
Thanks in advance for the help.
You are wrong about function $g$. You have, for each $x\in\Bbb R$,\begin{align}g(x)&=f\bigl(|x|\bigr)\\&=\left\{\begin{array}{l}|x|^2&\text{ if }|x|\geqslant0\\|x|&\text{ if }|x|<0\end{array}\right.\\&=x^2\end{align}since $|x|^2=x^2$ and you never have $|x|<0$.
So, $g=h$ (and yes, that's a typo in the statement of part b)).