2.17.2 Applications

tationary/Turning points:

A stationary/Turning point is a point on a curve or surface where the gradient, or derivative, of the function is zero.

• A stationary/Turning point can be either a maximum, a minimum, or a point of inflection.
  • A maximum or minimum point is a point where the function reaches a local maximum or minimum value, respectively.
    • If the second derivative at a stationary point is positive, it is a minimum point, and if the second derivative is negative, it is a maximum point.
  • A point of inflection is a point where the curvature of the function changes sign.
    • A point of inflection can be identified by looking for a stationary point where the second derivative is zero but the third derivative is nonzero.

To find coordinates of a turning point:

• Put the derivative of the function equal to zero.

\( \frac{dy}{dx}=0 \)

• In this way, you will find the \(x-coordinate\) of the turning point.
• Substitute \(x-coordinate\) into the equation of the graph to get \(y-coordinate\) of the turning point.

Tip:

Read the question carefully, sometimes only \(x-coordinate\) is required.

Worked example:

A curve has the equation  \(y=x^3+8x^2+5x\).

Work out the coordinates of the two turning points.

At turning points, \( \frac{dy}{dx}=0 \)

\( \frac{dy}{dx}=0 \)

\( \frac{d}{dx}(x^3+8x^2+5x)=0 \)

\(3x^2+16x+5=0\)

\(3x^2+15x+x+5=0\)

\(3x(x+5)+1(x+5)=0\)

\((3x+1)(x+5)=0\)

\(3x+1=0,\  x+5=0\)

Work out the corresponding  y-coordinates.

\(x=-\frac{1}{3},\  y=(-\frac{1}{3})^3+8(-\frac{1}{3}^2)+5(-\frac{1}{3})=-\frac{22}{27}\)

\(x=-5,\   y=(-5)^3+8(-5)^2+5(-5)=50\)

Turning points are \((-\frac{1}{3},\ -\frac{22}{27})\ and\ (-5,\ 50)\).

To classify which point is maximum and which point is minimum

• Find second derivative \(\frac{d^2x}{dy^2}\)
  • Derivative of \(\frac{dy}{dx}\)
  • or differentiate the function twice
• Substitute the x-coordinate of the stationary point in\(\frac{d^2y}{dx^2}\)
• to get a numerical value
  • If this value is negative, the stationary point is a maximum point
  • If this value is positive, the stationary point is a minimum point
  • If this value is zero, unfortunately the test has failed, go back and sketch the graph to check whether the point is maximum or minimum.

Worked example:

A curve has equation \(y=x^3-3x+4\).

Work out the coordinates of the two stationary points.

At stationary/turning points, \(\frac{dy}{dx} = 0\)

\(\frac{dy}{dx} = 0\)

\(\frac{d}{dx}\left( x^3 – 3x + 4 \right) = 0\)

\(3x^2 – 3 = 0\)

\(x^2 – 1 = 0\)

\(x^2 = 1\)

\(x = 1, \quad x = -1\)

Work out the corresponding \(y-coordinates\)

\(x = 1, \quad y = 1^3 – 3(1) + 4 = 2\)

\(x = -1, \quad y = (-1)^3 – 3(-1) + 4 = 6\)

Stationary points are \(1,\ 2\ and\ -1,\ 6.\)

Determine whether each stationary point is a maximum or a minimum.

Give reasons for your answers. 

Find second derivative

\(\frac{d^2 y}{dx^2} = \frac{d}{dx} (3x^2 – 3)\)

\(\frac{d^2 y}{dx^2} = 6x\)

Calculate the value of \(\frac{d^2 y}{dx^2}\) at each stationary point

\(x = -1, \quad \left. \frac{d^2 y}{dx^2} \right|_{x = -1} = 6(-1) = -6\)

\(-1,\ 6\) is a maximum point because \(\frac{d^2 y}{dx^2} < 0 \text{ at } x = -1\)

\(x = 1, \quad \left. \frac{d^2 y}{dx^2} \right|_{x = 1} = 6(1) = 6\)

\(1,\ 2\) is a minimum point because \(\frac{d^2 y}{dx^2} > 0 \text{ at } x = 1\) 

Test yourself

Question 1:

Differentiate

\(6+4x-x^2\) .

[2]

Find the coordinates of the turning point of the graph of

\(y=6+4x-x^2\)

[2]

Question 2:

Find the two stationary points on the graph of \(y=x^4-4x^3\).

[6]

Question 3:

The diagram shows a sketch of the curve  \(y=x^2+3x-4\).

• Differentiate \(y=x^2+3x-4\) 

[2]

• Find the equation of the tangent to the curve at the point 2, 6

[3]

Question 4:

\(y=x^p+2x^q\)

\(\frac{dy}{dx}=11x^{10}+10x^4\),  where \(\frac{dy}{dx}\) is the derived function.

Find the value of p and the value of q.

[2]

Question 5:

Calculate the gradient of \(y=24+5x-x^2\ at\ x=-1.5\) .

[3]