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Discuss the differentiability of the function π at π₯ is equal to one given that π of π₯ is equal to two π₯ plus eight if π₯ is less than one and π of π₯ is equal to π₯ squared plus nine if π₯ is greater than or equal to one.
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Weβre given a piecewise-defined function π of π₯.
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We need to determine whether the function π is differentiable when π₯ is equal to one.
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Thereβs a few different ways of doing this.
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For example, we could first check whether our function π of π₯ is continuous when π₯ is equal to one.
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Then if π of π₯ is continuous when π₯ is equal to one, we can check the slope from the left and from the right when π₯ is equal to one by differentiating both pieces of our function.
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If these two slopes match, then we can say π of π₯ is continuous when π₯ is equal to one.
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However, in this video, weβre going to determine this directly from the definition of a derivative.
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Letβs start by recalling what we mean by the derivative of π of π₯ with respect to π₯ at a point π₯ naught.
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This is equal to the limit as β approaches zero of π of π₯ naught plus β minus π of π₯ naught all divided by β.
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And this assumes this limit exists.
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If this limit does not exist, then we say that our function π of π₯ is not differentiable at the point π₯ naught.
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In this case, weβre given the piecewise function π of π₯.
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And we want to know if itβs differentiable at the point where π₯ is equal to one.
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So weβll set π₯ naught equal to one.
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We can then substitute π₯ naught is equal to one into our definition of the derivative.
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So determining whether π of π₯ is differentiable when π₯ is equal to one is the same is determining whether this limit exists.
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However, we canβt directly evaluate this limit because our function π of π₯ is given as a piecewise function and π₯ one is the endpoints of one of these intervals.
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So we need to remember a fact about limits.
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Instead of finding the limit as β approaches zero of this expression, weβll find the limit as β approaches zero from the right of this expression and the limit as β approaches zero from the left of this expression.
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If both of these limits exist and are equal, then we know our function π of π₯ is differentiable at this point.
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However, if either limit does not exist or these limits are not equal, then we know that our function is not differentiable when π₯ is equal to one.
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Letβs start by evaluating the limit as β approaches zero from the left of π of one plus β minus π of one all divided by β.
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To evaluate this limit, we first need to notice that β is approaching zero from the left.
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This means that all of our values of β will be less than zero.
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And if β is less than zero, then one plus β will be less than one.
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And we know from our piecewise definition of the function π of π₯ when π₯ is less than one, π of π₯ is equal to the linear function two π₯ plus eight.
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So in this case, to evaluate π at one plus β, we just substitute π₯ is equal to one plus β into the linear function two π₯ plus eight.
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This gives us two times one plus β add eight.
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We might be tempted to substitute one into this linear function.
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But remember, we need to use the piecewise definition of π of π₯.
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When π₯ is equal to one, π of π₯ is exactly equal to the quadratic π₯ squared plus nine.
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Substituting π₯ is equal to one into our quadratic gives us one squared plus nine.
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So we now have to evaluate the limit as β approaches zero from the left of two times one plus β plus eight minus one squared plus nine all divided by β.
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To do this, we need to simplify the expression inside of our limit.
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Weβll start with the first term in our numerator.
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Weβll distribute two over the parentheses.
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This gives us two plus two β plus eight.
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And of course, we can simplify this since two plus eight is equal to 10.
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Next, weβll evaluate the second term in our numerator. we have one squared plus nine is equal to one plus nine.
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And of course, we can simplify one plus nine to give us 10.
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So now, we have the limit as β approached zero from the left of two β plus 10 minus 10 all divided by β.
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Of course, we can keep simplifying this.
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In our numerator, 10 minus 10 is equal to zero.
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So in actual fact, this entire limit just simplified to give us the limit as β approaches zero from the left of two β divided by β.
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And of course, we know β is approaching zero from the left.
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In particular, this means that β is not equal to zero.
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Itβs just getting closer and closer to zero.
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So we can cancel the shared factor of β in our numerator and our denominator, meaning weβre just left with the limit as β approaches zero from the left of the constant two.
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But two is a constant, so this limit evaluates to give us two.
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Therefore, as π₯ approaches one from the left, the slope of our function π of π₯ is equal to two.
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We now need to do exactly the same thing from the right.
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So letβs clear some space and evaluate the limit as β approaches zero from the right of π of one plus β minus π of one all divided by β.
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This time, we have β is approaching zero from the right.
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This means we know all of our values of β will be bigger than zero.
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And if β is just bigger than zero, this means one plus β will be bigger than one.
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So once again, we need to use the piecewise definition of our function π of π₯ to evaluate π of one plus β.
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We know when π₯ is greater than or equal to one, π of π₯ is exactly equal to the quadratic π₯ squared plus nine.
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So because one plus β is greater than one, to evaluate π at one plus β, we substitute π₯ is equal to one plus β into this quadratic.
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This gives us one plus β all squared plus nine.
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We already found π evaluated at one in our previous limit.
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To do this, we use the piecewise definition of π of π₯.
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We just substitute π₯ is equal to one into the quadratic π₯ squared plus nine.
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This gave us one squared plus nine, which we simplified to give us 10.
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So we now need to evaluate the limit as β approaches zero from the right of one plus β all squared plus nine minus 10 all divided by β.
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To do this, we can start simplifying.
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First, we know that nine minus 10 is equal to negative one.
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Next, we want to distribute the square over our parentheses.
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Doing this by using the FOIL method or by using binomial expansion, we get one plus two β plus β squared.
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Then we need to subtract one from this and divide through by β.
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And in fact, we can see we can keep simplifying.
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In our numerator, we have one minus one, which simplifies to give us zero.
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So this gives us the limit as β approaches zero from the right of two β plus β squared all divided by β.
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And once again, this limit is as β is approaching zero from the right.
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β is getting closer and closer to zero, but β is never equal to zero.
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This means we can cancel the shared factor of β in our numerator and our denominator.
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This leaves us with the limit as β approaches zero from the right of two plus β.
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And of course, as β is approaching zero from the right, our value of β is approaching zero and our constant two remains constant.
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So this limit just evaluates to give us two.
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And now, we can see directly from the definition of the slope that these two slopes are equal.
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And because these two values are equal, we can conclude that π of π₯ is differentiable at π₯ is equal to one.
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Itβs also worth reiterating at this point, we could only use this method because we directly worked with the definition of differentiability at a point.
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If we had instead wanted to work with other rules of differentiation, such as the power rule for differentiation on our function π of π₯, we would also need to show that π of π₯ is continuous when π₯ is equal to one.
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However, as weβve seen, itβs not necessary to use this method.
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We can just do this directly from the definition of a derivative.
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Therefore, we were able to show the function π of π₯ is equal to two π₯ plus eight if π₯ is less than one and π of π₯ is equal to π₯ squared plus nine if π₯ is greater than or equal to one is differentiable at π₯ is equal to one.