Question 6.17: Find the turning points for the curve, y=−x³ + 9x² − 24x + 2......

The graph of y = −x³ + 9x² − 24x + 26 is drawn in Figure 6.16 to confirm and show the answers found using the following method.

Step 1: Find first and second derivatives,

y = −x³ + 9x² − 24x + 26

Slope: \frac {dy}{dx} = −3x² + 18x − 24

Second derivative : \frac {d^2y}{dx^2} = −6x + 18

Step 2: At turning points, slope is zero, therefore, solve the equation dy/dx = 0,

\frac {dy}{dx} = 0  at turning points

−3x² + 18x − 24 = 0

x² − 6x + 8 = 0        dividing both sides by −3

(x − 4)(x − 2) = 0

x = 4 or  x = 2

(If it is difficult to find the factors for a quadratic, use the ‘minus b’ method to find the roots.) Therefore, there are two turning points, one at the point whose x-coordinate is x = 4 and the other at the point whose x-coordinate is x = 2.

Step 2a: Find y.

Since we know the x-coordinate, find y from the equation of the curve:

y = −x³ + 9x² − 24x + 26
y = −(2)³ + 9(2)² − 24(2) + 26      substituting x = 2
= −8 + 36 − 48 + 26 = 6 ∴ y = 6  when x = 2
y = −(4)³ + 9(4)² − 24(4) + 26       substituting x = 4
= −64 + 144 − 96 + 26 = 10 ∴ y = 10 when x = 4

Step 3: Determine whether each turning point is a maximum or a minimum from either the ‘slope test’ or the ‘second-derivative test’. Both methods are illustrated below.

Turning point at x = 2
Method A: slope test

Evaluate slope: \frac {dy}{dx} = −3x² + 18x − 24 at any convenient value of x before x = 2, at x = 2 and after x = 2, the turning point

For example, take x = 0 as the point before x = 2 and x = 3 as the point after x = 2

The slope at each x-value is given below followed by a rough sketch:

Hence there is a minimum point at x = 2

Method B

Evaluate the second derivative at x = 2

\frac {d^2y}{dx^2} = −6x + 18

= −6(2) + 18 = 6

POSITIVE
Therefore, a minimum at x = 2 (2, 6) is a minimum point

Turning point at x = 4
Method A: slope test

Evaluate slope: \frac {dy}{dx} = −3x² + 18x − 24 at any convenient value of x before x = 4, at x = 4 and after x = 4, the turning point

For example, take x = 3 as the point before x = 4 and x = 5 as the point after it

The slope at each x-value is given below followed by a rough sketch:

Hence there is a maximum point at x = 2

Method B
Evaluate the second derivative at x = 4

\frac {d^2y}{dx^2} = −6x + 18

= −6(4) + 18 = −6

NEGATIVE

Therefore, (4, 10) is a maximum point