Calculus Examples

$e^{1 + \ln (x)}$

Step 1

Find the asymptotes.

Tap for more steps...

Step 1.1

Set the argument of the logarithm equal to zero.

$x = 0$

Step 1.2

The vertical asymptote occurs at $x = 0$ .

Vertical Asymptote: $x = 0$

Step 2

Find the point at $x = 1$ .

Tap for more steps...

Step 2.1

Replace the variable $x$ with $1$ in the expression.

$f (1) = e^{1 + \ln (1)}$

Step 2.2

Simplify the result.

Tap for more steps...

Step 2.2.1

The natural logarithm of $1$ is $0$ .

$f (1) = e^{1 + 0}$

Step 2.2.2

Add $1$ and $0$ .

$f (1) = e$

Step 2.2.3

The final answer is $e$ .

$e$

Step 2.3

Convert $e$ to decimal.

$y = 2.71828182$

Step 3

Find the point at $x = 2$ .

Tap for more steps...

Step 3.1

Replace the variable $x$ with $2$ in the expression.

$f (2) = e^{1 + \ln (2)}$

Step 3.2

The final answer is $e^{1 + \ln (2)}$ .

$e^{1 + \ln (2)}$

Step 3.3

Convert $e^{1 + \ln (2)}$ to decimal.

$y = 5.43656365$

Step 4

Find the point at $x = 3$ .

Tap for more steps...

Step 4.1

Replace the variable $x$ with $3$ in the expression.

$f (3) = e^{1 + \ln (3)}$

Step 4.2

The final answer is $e^{1 + \ln (3)}$ .

$e^{1 + \ln (3)}$

Step 4.3

Convert $e^{1 + \ln (3)}$ to decimal.

$y = 8.15484548$

Step 5

The log function can be graphed using the vertical asymptote at $x = 0$ and the points $(1, 2.71828182), (2, 5.43656365), (3, 8.15484548)$ .

Vertical Asymptote: $x = 0$

$\begin{array}{c} x & y \\ 1 & 2.718 \\ 2 & 5.437 \\ 3 & 8.155 \end{array}$

Step 6