Question

The function \( x^4 e^{-2x^2/3} \) (for \( x > 0 \)) has a maximum at a value of \( x \) equal to ...........
(Round off to two decimal places)

Hint:

To find the maximum of a function, differentiate and set the derivative equal to zero. Then solve for \( x \).

Answer 1

Correct Answer: 5.97 - 6.06

Given function: $f(x) = x^4 e^{-2x/3}$ for $x > 0$

To find maximum, take derivative and set equal to zero:

$$f'(x) = \frac{d}{dx}(x^4 e^{-2x/3})$$

Using product rule: $$f'(x) = 4x^3 e^{-2x/3} + x^4 \cdot e^{-2x/3} \cdot \left(-\frac{2}{3}\right)$$

$$f'(x) = 4x^3 e^{-2x/3} - \frac{2x^4}{3} e^{-2x/3}$$

$$f'(x) = e^{-2x/3} \left(4x^3 - \frac{2x^4}{3}\right)$$

$$f'(x) = e^{-2x/3} \cdot x^3 \left(4 - \frac{2x}{3}\right)$$

Set $f'(x) = 0$:

Since $e^{-2x/3} > 0$ and $x^3 > 0$ for $x > 0$:

$$4 - \frac{2x}{3} = 0$$

$$4 = \frac{2x}{3}$$

$$12 = 2x$$

$$x = 6$$

Verify it's a maximum (second derivative test or check sign change):

• For $x < 6$: $4 - \frac{2x}{3} > 0$, so $f'(x) > 0$ (increasing)
• For $x > 6$: $4 - \frac{2x}{3} < 0$, so $f'(x) < 0$ (decreasing)

Therefore, $x = 6$ is a maximum.

Answer: 6.00