Video: Finding the Value of an Unknown in a Polynomial Function Given the Value of Its First Derivative at a Point

Given that 𝑓(π‘₯) = βˆ’π‘₯Β² + π‘šπ‘₯ + 1, determine π‘š if 𝑓′(3) = 1.

03:15

Video Transcript

Given that 𝑓 of π‘₯ is equal to negative π‘₯ squared plus π‘šπ‘₯ plus one, determine π‘š if 𝑓 prime of three is equal to one.

Looking at the function we’ve been given, we can see that π‘š represents a missing coefficient in this polynomial 𝑓 of π‘₯. And we need to work out its value. We’re also given that 𝑓 prime of three is equal to one, which means the first derivative of our function 𝑓 evaluated when π‘₯ is equal to three is one.

To answer this question then, we’re going to need to find an expression for the first derivative of our function. And as 𝑓 of π‘₯ is a polynomial, we can do this by applying the power rule of differentiation. This tells us that, for real values of π‘Ž and 𝑛, the derivative with respect to π‘₯ of π‘Žπ‘₯ to the 𝑛th power is equal to π‘Žπ‘›π‘₯ to the power of 𝑛 minus one. We multiplied by the original exponent of 𝑛 and then reduced the exponent by one.

We also know that if we want to find the derivative of a sum or difference of terms or functions, then this is equal to the sum or difference of their individual derivatives, which essentially just means we can differentiate term by term and add their individual derivatives together. So let’s find a general expression for 𝑓 prime of π‘₯, the first derivative of this function. Applying the power rule of differentiation to the first term gives negative two π‘₯.

In order to differentiate the second term, we may find it helpful to think of π‘₯ as π‘₯ to the power of one. And then, applying the power rule of differentiation, we have that the derivative of π‘šπ‘₯ to the power of one will be equal to π‘š multiplied by one multiplied by π‘₯ to the power of zero. In the same way, we may find it helpful to think of the constant one as just one multiplied by π‘₯ to the power of zero. So then, applying the power rule of differentiation, we have one multiplied by zero multiplied by π‘₯ to the power of zero minus one.

We recall though that π‘₯ to the power of zero is one. And multiplying anything by zero gives zero. So, in fact, the derivative of our final term is just zero. And the derivative of our second term is just π‘š multiplied by one. Our expression for 𝑓 prime of π‘₯ therefore simplifies to negative two π‘₯ plus π‘š.

Now, we recall that we know that the first derivative evaluated when π‘₯ is equal to three is one. So substituting three for π‘₯ and one for 𝑓 prime of π‘₯ gives the equation one equals negative two multiplied by three plus π‘š. This simplifies to one equals negative six plus π‘š. And then, adding six to each side, we find that the solution to this equation is π‘š equals seven.

We have therefore completed the problem. We used the power rule of differentiation to find a general expression for the first derivative 𝑓 prime of π‘₯. And then, we used the fact that we know the first derivative when π‘₯ is equal to three is one to form an equation. We’ve hence found that the missing coefficient π‘š is equal to seven.

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