Video: APCALC03AB-P1B-Q40-494176086464

If 𝑓(π‘₯) is differential and 𝑓(π‘₯) = {𝑛π‘₯⁴ βˆ’ 3π‘₯Β², π‘₯ ≀ 2 and π‘šπ‘₯Β² + 2, π‘₯ > 2, what is the value of π‘š?

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Video Transcript

If 𝑓 of π‘₯ is differential and 𝑓of π‘₯ equals 𝑛π‘₯ to the fourth power minus three π‘₯ squared for π‘₯ less than or equal to two and π‘šπ‘₯ squared plus two for π‘₯ greater than two, what is the value of π‘š?

This unknown value π‘š is a coefficient in the definition of our function 𝑓 of π‘₯. So, it’s likely that we’re going to need to form some equations in order to find its value. Now, 𝑓 of π‘₯ is a piecewise function. It’s defined differently either side of π‘₯ equals two. However, we’re told in the question this function 𝑓 of π‘₯ is differentiable. What this tells us is that the left-hand limit of the derivative 𝑓 prime of π‘₯, as we approach π‘₯ equals two from below, is the same as the right-hand limit of the derivative 𝑓 prime of π‘₯, as we approach π‘₯ equals two from above.

Each part of our piecewise function 𝑓 of π‘₯ is just a polynomial in π‘₯. So, we can find the derivative of each part using the power rule of differentiation. The derivative, when π‘₯ is less than or equal to two, is four 𝑛π‘₯ cubed minus six π‘₯. And the derivative when π‘₯ is greater than two is two π‘šπ‘₯. Remember, the derivative of a constant in this case, positive two, is just zero.

Now, remember, if 𝑓 of π‘₯ is differentiable, then these two derivatives must be equal to one another at the point where the definition changes. That’s when π‘₯ is equal to two. So, by substituting π‘₯ equals two into each part of our derivative and setting the two expressions equal to one another, we can form an equation. For 𝑛 multiplied by two cubed minus six multiplied by two is equal to two π‘š multiplied by two. Which simplifies to 32𝑛 minus 12 equals four π‘š.

We can simplify it by dividing each of our coefficients by four and then rearrange the equation slightly so that we have eight 𝑛 and minus π‘š equals three. The trouble is, though, this is only one equation and we have two unknowns 𝑛 and π‘š. So, in order to find the values of 𝑛 and π‘š, we’re going to need another equation which we can solve simultaneously with this one, which we’ll call equation one.

The key information we need in order to form this second equation is that if a function is differentiable at a point, then it is also continuous at that point. So, if our function 𝑓 of π‘₯ is differentiable, it must also be continuous throughout its domain. This tells us that the left-hand limit of the function itself must be equal to the right-hand limit of the function itself when π‘₯ equals two.

In the same way then, we can substitute π‘₯ equals two into each part of the definition of the function itself and equate these two expressions to give 𝑛 multiplied by to the fourth power minus three multiplied by two squared is equal to π‘š multiplied by two squared plus two. That simplifies to 16𝑛 minus 12 equals four π‘š plus two. We can then rearrange the terms and divide each of our coefficients by two to give eight 𝑛 minus two π‘š equals seven. And we have our second equation.

What we now have is a pair of linear simultaneous equations in 𝑛 and π‘š, which we can solve simultaneously in order to find their values. We notice, first of all, that as the coefficients of 𝑛 are the same in both equations, we can eliminate 𝑛 by subtracting equation two from equation one. That leaves negative π‘š minus negative two π‘š, which is π‘š, is equal to three minus seven, which is negative four.

Now, there’s no need to determine the value of 𝑛, but if we did want to, we could substitute this value of π‘š back in to either of our two equations. So, by recalling that if a function is differentiable, it must also be continuous, and by equating left-hand and right-hand limits of both the function and its derivative when π‘₯ equals two, we’ve found that the value of π‘š is negative four.

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