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Let π be a function whose derivative is given by π prime of π₯ equals five π₯ over two π₯ squared plus six all squared.
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On which of the following intervals of π₯ is the graph of π concave up?
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Is it a) the open interval from zero to β.
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B) The open interval negative β to zero.
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Is it c) the union of the open interval from negative β to negative one and the open interval one to β?
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Or d) the open interval negative one to one.
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Weβre going to need to be a little bit careful here.
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We have not been given an expression for the function π itself, but one for its derivative, π prime of π₯.
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Weβre looking to find the intervals on which the graph of π is concave up.
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So what does this mean?
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When a graph is concave up, we say its first derivative, π prime of π₯, is increasing.
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We can consider the rate of change of that first derivative.
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Thatβs the derivative of the derivative, π double prime of π₯.
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And what this means is π double prime of π₯ must be greater than zero.
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So to answer this question, weβre going to need to differentiate π prime of π₯ to find the second derivative and then establish on which intervals of π₯ that is greater than zero.
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Now π prime of π₯ is the quotient of two functions.
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So we use the quotient rule.
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And this says that the derivative of the quotient of two differentiable functions, π’ and π£, is π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all over π£ squared.
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The numerator of our fraction is five π₯.
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So weβre going to let π’ be equal to five π₯.
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And π£ is the denominator.
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Itβs two π₯ squared plus six squared.
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It should be quite clear from our formula for the quotient rule that weβre going to need to differentiate each of these functions with respect to π₯.
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Well, the derivative of five π₯ is fairly straightforward.
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Itβs simply five.
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But what about the derivative of this composite function, two π₯ squared plus six all squared?
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Now we could use the chain rule.
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Or we can use a special case of the chain rule, which is called the general power rule.
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What we do is we multiply the entire inner function by the value of the exponent, so by two.
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And then we reduce that exponent by one.
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Then we multiply that by the derivative of the inner function.
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Well, the derivative of two π₯ squared plus six with respect to π₯ is four π₯.
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So we find that dπ£ by dπ₯ in its simplest form is eight π₯ times two π₯ squared plus six.
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Letβs substitute everything we have into the quotient rule.
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The derivative of π prime of π₯, which is π double prime of π₯, is π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all over π£ squared.
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We can simplify the denominator of this expression by multiplying the exponent.
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So we get two π₯ squared plus six to the fourth power.
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On our numerator, we can factor out two π₯ squared plus six.
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And so we find that π double prime of π₯ is two π₯ squared plus six times five times two π₯ squared plus six minus 40π₯ squared over two π₯ squared plus six to the fourth power.
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We distribute the first set of parentheses to give us 10π₯ squared plus 30.
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And so we see this simplifies to two π₯ squared plus six times 30 minus 30π₯ squared all over two π₯ squared plus six to the fourth power.
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Now we say that, for the graph to be concave up, the second derivative must be greater than zero.
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So we need to find the intervals of π₯ for which this is the case.
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And this might look quite complicated to start.
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However, we know that any real number to the fourth power is greater than zero.
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We know two π₯ squared is greater than zero.
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And therefore, two π₯ squared plus six is greater than zero.
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So we need to work out where the other part of the expression 30 minus 30π₯ squared is greater than zero.
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Then we have the product of two positives divided by a positive, which we know to be a positive.
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Letβs clear some space and solve this inequality.
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Weβll divide through by 30.
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And that gives us one minus π₯ squared is greater than zero.
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Letβs factor the expression one minus π₯ squared and for now just set it equal to zero.
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We get one minus π₯ times one plus π₯ equals zero.
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And for this to be true, either one minus π₯ must be equal to zero or one plus π₯ must be equal to zero.
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And so we see that when π₯ is equal to one and negative one, the expression one minus π₯ squared is equal to zero.
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So where is it greater than zero?
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Well, there are a number of ways we can achieve this.
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But one way is to draw the graph.
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Now we see the graph of π¦ equals one minus π₯ squared is greater than zero here.
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Thatβs values of π₯ greater than negative one and less than one.
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This means our second derivative is greater than zero for values of π₯ on the open interval negative one to one.
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And in turn, the graph of π is concave up over that same interval.
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So the correct answer is d) the open interval negative one to one.