How do you sketch y=e^absx?

1 Answer
Feb 4, 2017

See explanation...

Explanation:

When $x = 0$ the given formula gives $y = {e}^{\left\mid \textcolor{b l u e}{0} \right\mid} = 1$

Also note that since $y$ is a function of $\left\mid x \right\mid$, its graph will be symmetrical about the $y$ axis.

The function $f \left(x\right) = {e}^{x}$ is well known for being equal to its own derivative. That is, the instantaneous slope at any point is itself ${e}^{x}$.

In symbols we can write:

$\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$

So we find:

${\lim}_{x \to 0} \frac{d}{\mathrm{dx}} {e}^{x} = {\lim}_{x \to 0} {e}^{x} = {e}^{0} = 1$

Hence the slope of our given function approaches $1$ as $x \to {0}^{+}$, that is as $x \to 0$ from the right.

As a result, the graph of $y = {e}^{\left\mid x \right\mid}$ forms a V shape at $\left(0 , 1\right)$ with slope $\pm 1$

We can also evaluate $y = {e}^{\left\mid x \right\mid}$ for a few values of $x$ to provide some points through which the graph passes:

${e}^{\left\mid 1 \right\mid} \approx 2.718$

${e}^{\left\mid 2 \right\mid} \approx 7.389$

Putting this information together we can plot the graph of $y = {e}^{\left\mid x \right\mid}$ as a kind of steep U shape with a pointed vertex:
graph{(y-e^abs(x))(x^2+(y-1)^2-0.02)((x-1)^2+(y-e)^2-0.02)((x-2)^2+(y-e^2)^2-0.02) = 0 [-10.375, 9.625, -1.32, 8.68]}