A curve, C, has equation $$y = \frac{e^{3x}-5}{x^2}$$ Show that C has exactly one stationary point - AQA - A-Level Maths Mechanics - Question 13 - 2019 - Paper 1

Setting the derivative equal to zero gives:

$3xe^{3x} - 2(e^{3x} - 5) = 0$

Which simplifies to:

$e^{3x}(3x - 2) + 10 = 0$

This leads to:

$3x - 2 = 0$

Solving for x:

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

We need to check that this stationary point is valid by considering the denominator of the original equation, which is $x^2$. Since $x^2$ cannot be zero, we conclude that:

$x \neq 0$

Thus, $x = \frac{2}{3}$ is acceptable.

Lastly, we will analyze the behavior of the function around the stationary point to determine if it is indeed the only stationary point. The expression:

$\frac{dy}{dx} = \frac{e^{3x}(3x - 2) + 10}{x^3}$

shows that when $x < \frac{2}{3}$, $3x - 2 < 0$ leading to a positive derivative, and when $x > \frac{2}{3}$, $3x - 2 > 0$ leading to a negative derivative. This indicates that the curve has exactly one stationary point at $x = \frac{2}{3}$.