# Video: APCALC03AB-P1B-Q40-494176086464

If π(π₯) is differential and π(π₯) = {ππ₯β΄ β 3π₯Β², π₯ β€ 2 and ππ₯Β² + 2, π₯ > 2, what is the value of π?

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### Video Transcript

If π of π₯ is differential and πof π₯ equals ππ₯ to the fourth power minus three π₯ squared for π₯ less than or equal to two and ππ₯ squared plus two for π₯ greater than two, what is the value of π?

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

Each part of our piecewise function π 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 ππ₯ cubed minus six π₯. And the derivative when π₯ is greater than two is two ππ₯. Remember, the derivative of a constant in this case, positive two, is just zero.

Now, remember, if π of π₯ is differentiable, then these two derivatives must be equal to one another at the point where the definition changes. Thatβ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 π multiplied by two cubed minus six multiplied by two is equal to two π multiplied by two. Which simplifies to 32π minus 12 equals four π.

We can simplify it by dividing each of our coefficients by four and then rearrange the equation slightly so that we have eight π and minus π equals three. The trouble is, though, this is only one equation and we have two unknowns π and π. So, in order to find the values of π and π, weβre going to need another equation which we can solve simultaneously with this one, which weβ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 π 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 π multiplied by to the fourth power minus three multiplied by two squared is equal to π multiplied by two squared plus two. That simplifies to 16π minus 12 equals four π plus two. We can then rearrange the terms and divide each of our coefficients by two to give eight π minus two π equals seven. And we have our second equation.

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

Now, thereβs no need to determine the value of π, but if we did want to, we could substitute this value of π 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, weβve found that the value of π is negative four.