# How do you use the first and second derivatives to sketch y = e^(1-2x)?

Dec 15, 2017

See argument below.

#### Explanation:

$y = {e}^{1 - 2 x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} {e}^{1 - 2 x}$

Apply chain rule

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{1 - 2 x} \cdot \frac{d}{\mathrm{dx}} \left(1 - 2 x\right)$

Apply power rule

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 {e}^{1 - 2 x}$

Similarly, $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = + 4 {e}^{1 - 2 x}$

For extrema points of $y , \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

However, in this case, $- 2 {e}^{1 - 2 x} < 0 \forall x \in \mathbb{R}$

Consider, ${\lim}_{x \to + \infty} - 2 {e}^{1 - 2 x} = 0 \mathmr{and} {\lim}_{x \to + \infty} {e}^{1 - 2 x} = 0$

Also consider, $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = + 4 {e}^{1 - 2 x} > 0 \forall x \in \mathbb{R}$

Hence, it seems reasonable to deduce that $y$ tends to its minimum value of $0$ as $x$ tends to $+ \infty$

This helps us visualise the graph of $y$ below.

The other important point we need for the graph is at $x = 0$

Here, $y = {e}^{1 - 0} = e$

We see the point $\left(0 , e\right)$ on the graph below.

graph{e^(1-2x) [-5.69, 5.406, -1.722, 3.827]}