Find the turning points for the curve, y=−x³ + 9x² − 24x + 26.

Determine which point is a maximum and which is a minimum by using the second derivatives.

Step-by-Step

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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**

at x = 1 | at x = 2 | at x = 3 | |

dy/dx | -9 | 0 | 3 |

Direction | Negative | Zero | Positive |

at x = 3 | at x = 2 | at x = 5 | |

dy/dx | 3 | 0 | -9 |

Direction | Positive | Zero | Negative |

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