Sketching exponential curves

Stationary points, points of inflection and curve sketching for functions involving e^x, including asymptotes and finding minimum values.

Stationary points

Example 1: sketching y = xe⁻ˣ●●●

Sketch $y = xe^{-x}$, showing the stationary point and the point of inflection.

The curve passes through $(0, 0)$ only.

Intercepts

$y' = e^{-x} - xe^{-x} = (1 - x)e^{-x}$, which is zero at $x = 1$. The point is $\left(1, \tfrac1e\right)$.

First derivative

$y'' = -e^{-x} - (1 - x)e^{-x} = (x - 2)e^{-x}$. At $x = 1$, $y'' = -e^{-1} < 0$, so it is a maximum.

Second derivative

$y'' = 0$ at $x = 2$ and changes sign there, so $\left(2, \tfrac{2}{e^2}\right)$ is a point of inflection.

Inflection

As $x \to \infty$, $y \to 0$, so $y = 0$ is an asymptote on the right. As $x \to -\infty$, $y \to -\infty$.

Behaviour

y = xe⁻ˣ

Drag the point. The gradient is zero at x = 1, and the second derivative changes sign at x = 2.

Example 2●●

Find the minimum value of $y = e^x + e^{-x}$.

$y' = e^x - e^{-x} = 0$ gives $e^{2x} = 1$, so $x = 0$.

$y'' = e^x + e^{-x} > 0$, so it is a minimum. The minimum value is $y = 1 + 1 = 2$.

Example 3●●

Find the stationary point of $y = e^x - 4x$.

$y' = e^x - 4 = 0$ gives $x = \ln 4$.

$y = 4 - 4\ln 4$, and $y'' = e^x > 0$, so it is a minimum.

Flashcards

Solving $(ldots)e^{(ldots)} = 0$

Only the bracket can be zero, because $e^{(\ldots)} > 0$.

Maximum of $y = xe^{-x}$

$\left(1, \tfrac1e\right)$

As $x o infty$, $xe^{-x} o$

$0$

Practice