Question Video: Discussing the Monotonicity of a Polynomial Function | Nagwa Question Video: Discussing the Monotonicity of a Polynomial Function | Nagwa

# Question Video: Discussing the Monotonicity of a Polynomial Function Mathematics • Second Year of Secondary School

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Which of the following statements is true for the function π(π₯) = (βπ₯ β 6)Β³? [A] π(π₯) is increasing on β. [B] π(π₯) is decreasing on β. [C] π(π₯) is increasing on (ββ, β6) and decreasing on (β6, β). [D] π(π₯) is increasing on (β6, β) and decreasing on (ββ, β6).

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### Video Transcript

Which of the following statements is true for the function π of π₯ is equal to negative π₯ minus six all cubed? Is it option (A) π of π₯ is increasing on the set of real numbers? Option (B) π of π₯ is decreasing on the set of real numbers. Option (C) π of π₯ is increasing on the open interval from negative β to negative six and decreasing on the open interval from negative six to β. Or is it option (D) π of π₯ is increasing on the open interval from negative six to β and decreasing on the open interval from negative β to negative six?

In this question, weβre given four statements involving the monotonicity of a given function π of π₯. π of π₯ is equal to negative π₯ minus six all cubed. We need to determine which of these statements is true. To do this, we can start by recalling for a differentiable function π of π₯, we can determine the intervals which are decreasing and increasing by considering its derivative.

In particular, we recall if π is a differentiable function and if π prime of π₯ is greater than zero on an open interval from π to π, then we can conclude that π of π₯ is increasing on the open interval from π to π. Similarly, if π prime of π₯ is less than zero on an open interval from π to π, then π of π₯ is decreasing on the open interval from π to π. This is particularly useful because we can see that our function π of π₯ is differentiable for all real numbers. This is because weβre taking the cube of a linear function. So π of π₯ is a cubic polynomial. We can then use these results to determine the intervals on which π of π₯ is increasing and decreasing. We just need to determine an expression for π prime of π₯.

And thereβs a few different ways we can find π prime of π₯. For example, we could use the binomial formula to distribute our exponent over our parentheses. Or we could expand by using the FOIL method. Either case, we would end up with a cubic polynomial. We could differentiate this term by term by using the power rule for differentiation. This is not the only way we can differentiate this function. We can also notice that π of π₯ is the composition of two functions. We take a linear function and then we cube this value. And we can differentiate the composition of two functions by using the chain rule.

And in particular, because our outer function is a power function, we can differentiate this by using the general power rule, which is an application of the chain rule. This tells us the derivative of π of π₯ raised to the πth power is equal to π times π prime of π₯ multiplied by π of π₯ to the power of π minus one, where our value of π is a real constant and π of π₯ is a differentiable function. Weβll use this method to differentiate the function. However, itβs personal preference which one we want to use.

So weβll set our value of π equal to three and our inner function π of π₯ to be negative π₯ minus six. And itβs worth noting π of π₯ is a linear function. So π prime of π₯ will be the coefficient of π₯, which is negative one. We can now evaluate π prime of π₯ by substituting these values into the general power rule. π prime of π₯ is equal to three times negative one multiplied by negative π₯ minus six raised to the power of three minus one. We can then simplify this expression. π prime of π₯ is equal to negative three times negative π₯ minus six squared.

And now, we can notice something interesting. Negative π₯ minus six all squared is a square. This means itβs greater than or equal to zero for any value of π₯. In particular, thereβs only one input value of π₯ where this derivative will be equal to zero. Thatβs when π₯ is equal to negative six. We then multiply this by a negative constant. And remember, a negative multiplied by a positive is a negative. Therefore, we have shown that π prime of π₯ is less than or equal to zero for all values of π₯. And in particular, π prime of π₯ is equal to zero only when π₯ is equal to negative six.

We can then use our property. The derivative of π of π₯ is negative for all values of π₯ except when π₯ is negative six. This then tells us the function is decreasing on the open interval from negative β to negative six and the open interval from negative six to β. However, since our function is defined at negative six and its derivative is zero only at negative six, we can also say itβs decreasing at this point. In other words, weβve shown the function π of π₯ is equal to negative π₯ minus six all cubed is decreasing for all real values. This is option (B).

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