Video: Finding the Intervals of Increasing and Decreasing of a Quadratic Function

Determine the intervals on which the function 𝑓(π‘₯) = (βˆ’3π‘₯ βˆ’ 12)Β² is increasing and on which it is decreasing.

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Video Transcript

Determine the intervals on which the function 𝑓 of π‘₯ equals negative three π‘₯ minus 12 squared is increasing and on which it is decreasing.

Any function 𝑓 of π‘₯ is increasing if its derivative 𝑓 prime of π‘₯ is greater than zero. In the same way, the function is said to be decreasing if 𝑓 prime of π‘₯ is less than zero. Let’s firstly consider our function in this question negative three π‘₯ minus 12 all squared. We could differentiate this function immediately using the chain rule. Alternatively, we could distribute the parentheses in the function first.

Negative three π‘₯ multiplied by negative three π‘₯ is nine π‘₯ squared as multiplying two negatives gives us a positive. Negative three π‘₯ multiplied by negative 12 is equal to 36π‘₯. Likewise, negative 12 multiplied by negative three π‘₯ is 36π‘₯. Finally, negative 12 multiplied by negative 12 is 144. This means that the function 𝑓 of π‘₯ simplifies to nine π‘₯ squared plus 72π‘₯ plus 144. We can now differentiate each of the terms to work out an expression for 𝑓 prime of π‘₯.

Nine π‘₯ squared differentiated is equal to 18π‘₯ as two multiplied by nine is 18, and we decrease the exponent by one. 72π‘₯ differentiates to 72. Finally, any constant differentiates to zero, so the 144 disappears. The function 𝑓 prime of π‘₯ is equal to 18π‘₯ plus 72. The original function 𝑓 of π‘₯ was a quadratic function and 𝑓 prime of π‘₯ was a linear function. We can, therefore, see that the function will be increasing to the right of the minimum point and will be decreasing to the left of the minimum point.

This minimum value occurs when 𝑓 prime of π‘₯ is equal to zero. We can, therefore, set 18π‘₯ plus 72 equal to zero. Subtracting 72 from both sides of the equation gives us 18π‘₯ is equal to negative 72. Dividing both sides of this equation by 18 gives us π‘₯ is equal to negative four. The minimum point of the function 𝑓 of π‘₯ occurs when π‘₯ is equal to negative four.

This means that the function is increasing when π‘₯ is greater than negative four. The function must, therefore, be decreasing when π‘₯ is less than negative four. We were asked in the question to determine the intervals. The function is, therefore, increasing on the interval from negative four up to ∞ and it’s decreasing on the interval negative ∞ to negative four.

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