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Find the values of π and π given the function π is differentiable at π₯ is equal to one where π of π₯ is equal to negative π₯ squared plus four if π₯ is less than or equal to one and π of π₯ is equal to negative two ππ₯ minus π if π₯ is greater than one.
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Weβre given a piecewise-defined function π of π₯.
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And weβre told that this function π is differentiable when π₯ is equal to one.
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We need to use this to determine the values of π and π.
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The first thing we notice about this is when π₯ is equal to one, we can see weβre at the endpoints of our interval of the piecewise-defined function.
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In other words, when π₯ is equal to one, our function π of π₯ changes from being equal to negative π₯ squared plus four to being equal to negative two ππ₯ minus π.
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And at this point, thereβs a few different methods we could use to try and answer this question.
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For example, we might be tempted to directly use the definition of π being differentiable at π₯ is equal to one.
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And this would work.
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However, because our function π of π₯ is defined piecewise and π₯ is equal to one is one of the endpoints of this interval, we can actually do this in a simpler way.
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First, we recall if a function is differentiable at some point, then it must also be continuous at this point.
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In other words, since we know π is differentiable at π₯ is equal to one, we know that π must also be continuous when π₯ is equal to one.
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And we can see something interesting about our function π of π₯.
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We can see both pieces of this function are polynomials.
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And we know polynomials are continuous for all real values of π₯.
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So our function π of π₯ is piecewise continuous.
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And for a piecewise continuous function to be continuous at its endpoints, its endpoints must match.
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In other words, we know the limit as approaches one from the left of π of π₯ must be equal to the limit as π₯ approaches one from the right of π of π₯.
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Now, we could evaluate this limit directly.
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However, we need to remember π of π₯ is a piecewise continuous function.
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And because each piece is continuous, we can evaluate each of these limits by using direct substitution.
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So we just substitute π₯ is equal to one into negative π₯ squared plus four to evaluate the limit as π₯ approaches one from the left of π of π₯.
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We get negative one squared plus four.
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And we can do the same to evaluate the limit as π₯ approaches one from the right.
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We have negative two ππ₯ minus π is continuous.
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So we can evaluate this limit by using direct substitution.
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We just substitute π₯ is equal to one into negative two ππ₯ minus π.
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This gives us negative two π times one minus π.
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And because we know π is continuous, we know these two limits have to be equal.
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Letβs now simplify both sides of this equation.
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First, negative one squared plus four is equal to three.
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And we can simplify the right-hand side of this equation to give us negative two π minus π.
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But this is only one equation with two variables.
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So we need more information to find the values of π and π.
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To do this, weβll want to specifically use the fact that π is differentiable at π₯ is equal to one.
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And we know several different ways of explaining that π is differentiable at π₯ is equal to one.
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But one of these is a lot easier to work with for our function π of π₯.
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We can see that both of the parts of π of π₯ are defined as polynomials.
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And we already know how to differentiate polynomials term by term by using the power rule for differentiation.
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So instead of directly applying the definition of a derivative to our function π of π₯, we can instead look at the slope as π₯ approaches one from the left of π of π₯ and look at the slope as π₯ approaches one from the right of π of π₯.
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In other words, we know if π is differentiable at π₯ is equal to one, then the slope as π₯ approaches one from the left of π of π₯ must be equal to the slope as π approaches one from the right of π of π₯.
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And we use this because we can easily find an expression for π prime of π₯ when π₯ is less than one and when π₯ is greater than one.
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We just need to differentiate each piece of π of π₯ separately.
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We get π prime of π₯ is equal to the derivative of negative π₯ squared plus four with respect to π₯ if π₯ is less than one and π prime of π₯ is equal to the derivative of negative two ππ₯ minus π with respect to π₯ if π₯ is greater than one.
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And itβs worth reiterating at this point weβre not stating what π prime of π₯ is equal to when π₯ is equal to one.
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Weβre just finding an expression for the slope for all of the other values of π₯.
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Now, we can evaluate both of these derivatives by using the power rule for differentiation.
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We want to multiply by our exponents of π₯ and reduce this exponent by one.
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First, the derivative of negative π₯ squared plus four with respect to π₯ is equal to negative two π₯.
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Next, to differentiate our second function, we could again use the power rule for differentiation term by term.
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However, this is also a linear function.
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So we could just differentiate this by taking the coefficient of π₯, which is negative two π.
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This gives us π prime of π₯ is equal to negative two π₯ if π₯ is less than one and π prime of π₯ is equal to negative two π if π₯ is greater than one.
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Now, we can evaluate the slope as π₯ approaches one from the left of π of π₯ and the slope as π₯ approaches one from the right of π of π₯.
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First, when π₯ approaches one from the left, we can see π prime of π₯ is equal to negative two π₯.
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And of course, negative two π₯ is a continuous function.
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So we can evaluate this by using direct substitution.
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We just substitute π₯ is equal to one.
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We get negative two times one, which is equal to negative two.
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And we can do exactly the same as π₯ approaches one from the right.
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This time, π prime of π₯ will be equal to negative two π.
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But this time, negative two π is a constant.
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So this is just equal to negative two π.
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And remember, weβre told that π of π₯ is differentiable when π₯ is equal to one.
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So the slope as π₯ approaches one from the left of π of π₯ must be equal to the slope as π₯ approaches one from the right of π of π₯.
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In other words, we can equate these two values.
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We get negative two must be equal to negative two π.
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And if we divide both sides of this equation through by negative two, we see that π must be equal to one.
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Now, to find our value π, we substitute π is equal to one into our equation three is equal to negative two π minus π.
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Substituting π is equal to one gives us that three is equal to negative two times one minus π.
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And by simplifying and rearranging this equation, we can see that π must be equal to negative five.
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Therefore, if the function π of π₯ is equal to negative π₯ squared plus four if π₯ is less than or equal to one and π of π₯ is equal to negative two ππ₯ minus π if π₯ is greater than one is differentiable when π₯ is equal to one, then weβve shown that π must be equal to one and π must be equal to negative five.