Question Video: Finding Where the First Derivative of a Polynomial Function Equals a Given Value Mathematics • Higher Education

If the function 𝑓(π‘₯) = (1/3 π‘₯)Β³ + 2π‘₯Β² βˆ’ 36π‘₯ βˆ’ 24, find the values of π‘₯ that make 𝑓′(π‘₯) = βˆ’4.

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Video Transcript

If the function 𝑓 of π‘₯ is equal to one-third π‘₯ cubed plus two π‘₯ squared minus 36π‘₯ minus 24, find the values of π‘₯ that make 𝑓 prime of π‘₯ equal to negative four.

We’re given a function 𝑓 of π‘₯ which is a cubic polynomial. We need to determine the values of π‘₯ which make 𝑓 prime of π‘₯ equal to negative four. First, we need to recall what we mean by 𝑓 prime of π‘₯. Since 𝑓 is a function of π‘₯, this means the derivative of 𝑓 with respect to π‘₯. So we need to differentiate our function 𝑓 of π‘₯ with respect to π‘₯. We can see 𝑓 of π‘₯ is a cubic polynomial, and we know how to differentiate cubic polynomials. We’ll do this term my term by using the power rule for differentiation.

We recall the power rule for differentiation tells us for any real constants π‘Ž and 𝑛, the derivative of π‘Ž times π‘₯ to the 𝑛th power with respect to π‘₯ is equal to 𝑛 times π‘Ž times π‘₯ to the power of 𝑛 minus one. We multiply by the exponent of π‘₯ and then reduce this exponent by one. We want to apply this term by term on our function 𝑓 of π‘₯. Let’s start with the first term, our exponent of π‘₯ is equal to three.

So we want to multiply by this exponent of three and then reduce this exponent by one. This gives us three times one-third multiplied by π‘₯ to the power of three minus one. And of course, we can simplify this. Three multiplied by one-third is equal to one, and three minus one in our exponent simplifies to give us two. So the derivative of our first term simplified to give us π‘₯ squared. Let’s now move on to our second term. We can see the exponent of π‘₯ is equal to two.

Once again, we want to multiply by our exponent of two and then reduce this exponent by one. This gives us two times two π‘₯ to the power of two minus one. And this simplifies to give us four times π‘₯ to the first power. And of course, π‘₯ to the first power can be simplified to give us π‘₯. So the derivative of our second term simplifies to give us four π‘₯. We now want to move on to differentiate in our third term, negative 36π‘₯. However, we can see this is not written in the form π‘Ž times π‘₯ to the 𝑛th power. But we can write it in this form.

We need to remember π‘₯ to the first power is equal to π‘₯. So we can write this as negative 36 times π‘₯ to the first power. And now we can differentiate this by using the power rule for differentiation. We multiply by our exponent of one and then reduce this exponent by one. This gives us negative one times 36π‘₯ to the power of one minus one. And we can simplify this entire expression to negative 36π‘₯ to the zeroth power. But remember, any number raised to the zeroth power is equal to one. So we can simplify this to just be negative 36.

Finally, we want to find the derivative of the constant negative 24 with respect to π‘₯. We could do this by writing this as negative 24 times π‘₯ to the zeroth power because, remember, π‘₯ to the zeroth power is equal to one. However, it’s easier to do this by remembering since negative 24 is a constant, it does not vary as π‘₯ varies. So its derivative with respect to π‘₯ will be equal to zero. Therefore, we’ve shown that 𝑓 prime of π‘₯ is equal to π‘₯ squared plus four π‘₯ minus 36, but we’re not done. Remember, we need to find the values of π‘₯ that make 𝑓 prime of π‘₯ equal to negative four. So we should solve this equation equal to negative four.

So we want to solve π‘₯ squared plus four π‘₯ minus 36 is equal to negative four. We’ll do this by adding four to both sides of our equation. This gives us π‘₯ squared plus four π‘₯ minus 32 is equal to zero. And there’s a few different ways of solving this. For example, we could use a quadratic solver or the quadratic formula. We could also factor this by inspection by noticing eight multiplied by negative four is equal to negative 32 and eight plus negative four is equal to four.

So we’ve shown for 𝑓 prime of π‘₯ to be equal to negative four, π‘₯ plus eight times π‘₯ minus four must be equal to zero. And if the product of two factors is equal to zero, one of these factors must be equal to zero. Therefore, however, π‘₯ plus eight is equal to zero or π‘₯ minus four is equal to zero. And we can solve both of these equations, either π‘₯ is equal to negative eight or π‘₯ is equal to four. And since we found an expression for 𝑓 prime of π‘₯, we can verify both of these results. We’ll start by substituting π‘₯ is equal to negative eight into our expression for 𝑓 prime of π‘₯.

We get 𝑓 prime of negative eight is equal to negative eight all squared plus four times negative eight minus 36. And if we evaluate this expression, we see that we get negative four. And we can do the same when π‘₯ is equal to four. We get that 𝑓 prime of four is equal to four squared plus four times four minus 36. And if we evaluate this expression, we see we also get negative four as we expected.

Therefore, we’ve shown if the function 𝑓 of π‘₯ is equal to one-third π‘₯ cubed plus two π‘₯ squared minus 36π‘₯ minus 24 and 𝑓 prime of π‘₯ is equal to negative four, then however π‘₯ is equal to four or π‘₯ is equal to negative eight.

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