Stationary points of y=(3x+2)⁻¹+3x

Full worked solution

1. (a) Compute first derivative $\frac{dy}{dx}$

   We differentiate each term of $y$ separately, using the chain rule for the composite term $(3x+2{)}^{-1}$. The chain rule for $f(x)=(ax+b{)}^n$ gives $f^{\prime }(x)=na(ax+b{)}^{n-1}$; here $a=3$, $b=2$, $n=-1$. For the linear term, the derivative of $3x$ is $3$.

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

   Combine the two terms to get the first derivative:

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

   Therefore the first derivative is $\frac{dy}{dx}=3-\frac{3}{(3x+2{)}^2}$.

   Note.

   • Remember to multiply by the derivative of the inner function $3x+2$ for the chain rule term.
   • You may leave the derivative in fractional or negative-index form; both are acceptable.

2. (b) Compute second derivative $\frac{d^{2}y}{d{x}^2}$

   We differentiate the first derivative $\frac{dy}{dx}=-3(3x+2{)}^{-2}+3$ again, applying the chain rule to the first term. The constant term $3$ differentiates to 0. For the composite term, $n=-2$, $a=3$:

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

   The second derivative is therefore:

   $$\frac{d^2y}{dx^2} = 18(3x+2)^{-3} = \frac{18}{(3x+2)^3}$$

   Note.

   • The negative index becomes more negative when differentiated: $-2\to -3$.
   • The two negative signs multiply to a positive value, so the second derivative is positive when $(3x+2)>0$.

3. (c) Find $x$-coordinates of stationary points

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

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

   Rearrange terms to isolate the composite part:

   $$\frac{3}{(3x+2)^2} = 3$$

   Divide both sides by 3, then take reciprocals:

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

   Take the square root of both sides, accounting for positive and negative roots:

   $$3x + 2 = \pm 1$$

   Solve each case separately:

   • Case 1: $3x+2=1⟹3x=-1⟹x=-\frac{1}{3}$
   • Case 2: $3x+2=-1⟹3x=-3⟹x=-1$

   The $x$-coordinates of the stationary points are $x=-1$ and $x=-\frac{1}{3}$.

   Note.

   • Do not discard the negative root when taking the square root of 1; it gives a valid stationary point.
   • Note $x=-\frac{2}{3}$ is excluded from the domain of the curve, and is not a stationary point.

4. (d) Classify stationary points using the second derivative test

   We substitute each stationary point $x$-coordinate into the second derivative $\frac{d^{2}y}{d{x}^2}=\frac{18}{(3x+2{)}^3}$. The sign of the result tells us the nature of the point: positive = local minimum, negative = local maximum, zero = possible inflection point.

   For $x=-1$:

   $$\frac{d^2y}{dx^2}\bigg|_{x=-1} = \frac{18}{(3(-1)+2)^3} = \frac{18}{(-1)^3} = -18 < 0$$

   A negative second derivative means this point is a local maximum.

   For $x=-\frac{1}{3}$:

   $$\frac{d^2y}{dx^2}\bigg|_{x=-\frac{1}{3}} = \frac{18}{\left(3\left(-\frac{1}{3}\right)+2\right)^3} = \frac{18}{(1)^3} = 18 > 0$$

   A positive second derivative means this point is a local minimum.

   Therefore the stationary point at $x=-1$ is a local maximum, and the stationary point at $x=-\frac{1}{3}$ is a local minimum.

   Note.

   • If the second derivative were zero, you would need to test the sign of the first derivative around the point to classify it.
   • You do not need to compute the $y$-coordinate of the points for the second derivative test.

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