Stationary points of y=12x+(3x−2)⁻¹

Full worked solution

1. (a) Compute first derivative

   To find stationary points, we first calculate the first derivative of $y$ with respect to $x$ using the chain rule for the second term. For $(3x-2{)}^{-1}$, the derivative is $-1\cdot 3\cdot (3x-2{)}^{-2}=-3(3x-2{)}^{-2}$.

   $$\frac{dy}{dx} = \frac{d}{dx}\left[12x\right] + \frac{d}{dx}\left[(3x - 2)^{-1}\right] = 12 - 3(3x - 2)^{-2}$$

   Note.

   Recall the chain rule: $\frac{d}{dx}[f(g(x))]=f^{\prime }(g(x))\cdot g^{\prime }(x)$. Do not forget to multiply by the derivative of the inside function $3x-2$.

2. (b) Solve for stationary points (set $\frac{dy}{dx}=0$)

   Stationary points occur where $\frac{dy}{dx}=0$. Set the expression above equal to zero and rearrange to solve for $x$:

   $$12 - \frac{3}{(3x - 2)^2} = 0 \implies 12 = \frac{3}{(3x - 2)^2}$$

   Divide both sides by 3:

   $$4 = \frac{1}{(3x - 2)^2} \implies (3x - 2)^2 = \frac{1}{4}$$

   Take square roots of both sides:

   $$3x - 2 = \pm \frac{1}{2}$$

   Solve for $x$ for both cases:

   1. $3x-2=\frac{1}{2}⟹3x=\frac{5}{2}⟹x=\frac{5}{6}$
   2. $3x-2=-\frac{1}{2}⟹3x=\frac{3}{2}⟹x=\frac{1}{2}$

   The stationary points are at $x=\frac{1}{2}$ and $x=\frac{5}{6}$.

   Note.

   When taking square roots, always include both positive and negative roots to avoid missing one stationary point.

3. (c) Compute second derivative for classification

   To determine the nature of each stationary point, we calculate the second derivative by differentiating $\frac{dy}{dx}$ again, using the chain rule on the $-3(3x-2{)}^{-2}$ term:

   $$\frac{d^2y}{dx^2} = \frac{d}{dx}\left[12 - 3(3x - 2)^{-2}\right] = (-3)(-2)(3)(3x - 2)^{-3} = 18(3x - 2)^{-3} = \frac{18}{(3x - 2)^3}$$

   Note.

   The sign of the second derivative tells us the nature: $\frac{d^{2}y}{d{x}^2}>0$ = local minimum, $\frac{d^{2}y}{d{x}^2}<0$ = local maximum.

4. (d) Classify each stationary point

   Evaluate the second derivative at each stationary $x$-value:

   1. For $x=\frac{1}{2}$:

   $$3x - 2 = 3\left(\frac{1}{2}\right) - 2 = -\frac{1}{2} \implies \frac{d^2y}{dx^2} = \frac{18}{\left(-\frac{1}{2}\right)^3} = \frac{18}{-\frac{1}{8}} = -144$$

   Since $\frac{d^2y}{dx^2} < 0$, the point at $x = \frac{1}{2}$ is a local maximum.
   

   1. For $x=\frac{5}{6}$:

   $$3x - 2 = 3\left(\frac{5}{6}\right) - 2 = \frac{1}{2} \implies \frac{d^2y}{dx^2} = \frac{18}{\left(\frac{1}{2}\right)^3} = \frac{18}{\frac{1}{8}} = 144$$

   Since $\frac{d^2y}{dx^2} > 0$, the point at $x = \frac{5}{6}$ is a local minimum.
   

   Note.

   Do not substitute $x=\frac{2}{3}$ (where the function has a vertical asymptote) when testing values around stationary points, as the function is undefined there.

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