Stationary points questions with answers

Questions

1. Question 1Where dy/dx = 0

   Find the coordinates of the stationary point of the curve $y = x^{2} - 4x + 1$.

   Show the worked solution

   1. Step 1: Differentiate$\dfrac{dy}{dx} = 2x - 4(1)$
   2. Step 2: At a stationary point, dy/dx = 0$2x - 4 = 0$$x = 2$
   3. Step 3: Find y: put that value of x into the equation of the curve$y = (2)^{2} - 4(2) + 1$$= -3$
2. Question 2Where dy/dx = 0

   Find the coordinates of the stationary point of the curve $y = x + \dfrac{25}{x}$, where $x > 0$.

   Show the worked solution

   1. Step 1: Write every term as a power of x$y = x + 25x^{-1}$
   2. Step 2: Differentiate$\dfrac{dy}{dx} = 1 + 25(-1)x^{-2}$
   3. Step 3: At a stationary point, dy/dx = 0: rearrange for x², then take the positive root (x > 0)$1 - \dfrac{25}{x^{2}} = 0$$x^{2} = 25$$x = 5$
   4. Step 4: Find y: put that value of x into the equation of the curve$y = (5) + 25(5)^{-1}$$= 10$
3. Question 3Maximum or minimum?

   The curve $y = -x^{3} + 9x^{2} - 24x - 1$ has a stationary point at $x = 2$. Determine whether it is a maximum or a minimum point.

   Show the worked solution

   1. Step 1: Differentiate$\dfrac{dy}{dx} = -3x^{2} + 9(2)x - 24(1)$$= -3x^{2} + 18x - 24$
   2. Step 2: Differentiate again$\dfrac{d^{2}y}{dx^{2}} = -3(2)x + 18(1)$$= -6x + 18$
   3. Step 3: Put the stationary point's x into the second derivative$\dfrac{d^{2}y}{dx^{2}} = -6(2) + 18$$= 6$
   4. Step 4: Second derivative test: use the sign of the second derivative$\dfrac{d^{2}y}{dx^{2}} > 0$, so it is a minimum point.
4. Question 4Maximum or minimum?

   The curve $y = -x^{3} + 6x^{2} - 9x + 3$ has a stationary point at $x = 3$. Determine whether it is a maximum or a minimum point.

   Show the worked solution

   1. Step 1: Differentiate$\dfrac{dy}{dx} = -3x^{2} + 6(2)x - 9(1)$$= -3x^{2} + 12x - 9$
   2. Step 2: Differentiate again$\dfrac{d^{2}y}{dx^{2}} = -3(2)x + 12(1)$$= -6x + 12$
   3. Step 3: Put the stationary point's x into the second derivative$\dfrac{d^{2}y}{dx^{2}} = -6(3) + 12$$= -6$
   4. Step 4: Second derivative test: use the sign of the second derivative$\dfrac{d^{2}y}{dx^{2}} < 0$, so it is a maximum point.
5. Question 5The maximum or minimum point of a cubic

   Find the coordinates of the minimum point of the curve $y = x^{3} - 3x^{2} - 9x + 5$.

   Show the worked solution

   1. Step 1: Differentiate$\dfrac{dy}{dx} = 3x^{2} - 3(2)x - 9(1)$$= 3x^{2} - 6x - 9$
   2. Step 2: At a stationary point, dy/dx = 0. Simplify and factorise$3x^{2} - 6x - 9 = 0$$x^{2} - 2x - 3 = 0$$(x + 1)(x - 3) = 0$$x = -1$ or $x = 3$
   3. Step 3: Differentiate again$\dfrac{d^{2}y}{dx^{2}} = 3(2)x - 6(1)$$= 6x - 6$
   4. Step 4: Second derivative test at each stationary point: use the sign of the second derivativeWhen $x = -1$, $\dfrac{d^{2}y}{dx^{2}} = -12$When $x = 3$, $\dfrac{d^{2}y}{dx^{2}} = 12$$\text{The minimum point is at } x = 3.$
   5. Step 5: Find y: put that value of x into the equation of the curve$y = (3)^{3} - 3(3)^{2} - 9(3) + 5$$= -22$
6. Question 6The maximum or minimum point of a cubic

   Find the coordinates of the maximum point of the curve $y = x^{3} - 3x^{2} - 9x + 5$.

   Show the worked solution

   1. Step 1: Differentiate$\dfrac{dy}{dx} = 3x^{2} - 3(2)x - 9(1)$$= 3x^{2} - 6x - 9$
   2. Step 2: At a stationary point, dy/dx = 0. Simplify and factorise$3x^{2} - 6x - 9 = 0$$x^{2} - 2x - 3 = 0$$(x + 1)(x - 3) = 0$$x = -1$ or $x = 3$
   3. Step 3: Differentiate again$\dfrac{d^{2}y}{dx^{2}} = 3(2)x - 6(1)$$= 6x - 6$
   4. Step 4: Second derivative test at each stationary point: use the sign of the second derivativeWhen $x = -1$, $\dfrac{d^{2}y}{dx^{2}} = -12$When $x = 3$, $\dfrac{d^{2}y}{dx^{2}} = 12$$\text{The maximum point is at } x = -1.$
   5. Step 5: Find y: put that value of x into the equation of the curve$y = (-1)^{3} - 3(-1)^{2} - 9(-1) + 5$$= 10$