Find the value(s) of x for which y = [x(x − 2)]^{2} is an increasing function.

Advertisement Remove all ads

#### Solution

f(x)=[x(x−2)]^{2}

f'(x)=2[x(x−2)]{x−2+x}

f'(x)=4x(x−2)(x−1)

At critical point, f'(x)=0

4x(x−2)(x−1)=0

⇒x=0,1,2

Interval | f'(x)=4x(x−1)(x−2) |
Result |

(−∞,0) | f'(−1)=4(−1)(−2)(−3)=−24<0 | Decreasing |

(0,1) | f'(1/2)=4(1/2)(−1/2)(−3/2)=3/2>0 | Increasing |

(1,2) | f'(3/2)=4(3/2)(1/2)(−1/2)=−3/2<0 | Decreasing |

(2,∞) | f'(3)=4(3)(2)(1)=24>0 | Increasing |

So, the function is increasing in the interval (0,1)∪(2,∞).

Concept: Increasing and Decreasing Functions

Is there an error in this question or solution?

#### APPEARS IN

Advertisement Remove all ads