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Determine the intervals on which the function π of π₯ equals negative three π₯ plus the square root of nine π₯ squared plus one is concave up and down.
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Letβs begin by recalling what it means for a function to be either concave up or down.
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When a function is concave up, the tangents to the graph of the function lie below the graph itself.
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We see that the slope of these tangents is increasing.
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In the sketch, the slope is changing from negative to zero to positive.
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And so, the value of the slope is getting larger.
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We can say then that when a function is concave up, its first derivative π prime of π₯ is increasing.
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If π prime of π₯ is increasing, then its own derivative π double prime of π₯ must be positive.
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So, we can also conclude that when a function is concave up, its second derivative is greater than zero.
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When a function is concave down however, the opposite is true.
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The tangents to the graph lie above the graph itself.
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The first derivative π prime of π₯ is decreasing.
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In the sketch, we see itβs changing from positive to zero to negative.
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And so, the second derivative π double prime of π₯ will be negative.
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To answer this question then, weβre going to need to find an expression for the second derivative of our function π of π₯.
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So, weβre going to need to differentiate twice.
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Before we do though, we may find it easier to write the second part of the definition of π of π₯, thatβs the square root of nine π₯ squared plus one, as nine π₯ squared plus one to the power of one-half.
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Now, we can differentiate to find an expression for the first derivative π prime of π₯.
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Using the power rule of differentiation, the derivative of negative three π₯ is negative three.
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But what about the derivative of nine π₯ squared plus one to the power of one-half?
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Well, this is where we need to use the chain rule extension to the power rule, which tells us that if we have some function π of π₯ to the power of π, and we want to find its derivative with respect to π₯.
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Then, this is equal to π multiplied by π prime of π₯ multiplied by π of π₯ to the power of π minus one.
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We multiply by the exponent and reduce the exponent by one.
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But we also multiply by the derivative of our function π of π₯.
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So, differentiating nine π₯ squared plus one to the power of one-half then.
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We have π, thatβs the exponent, which is one-half.
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Then, the derivative of nine π₯ squared plus one, which is 18π₯.
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And then nine π₯ square plus one to the power of π minus one, which will be negative one-half.
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This all simplifies to negative three plus nine π₯ multiplied by nine π₯ squared plus one to the power of negative one-half.
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So, we have our expression for the first derivative π prime of π₯.
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But now, we need to differentiate again.
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The derivative of the first term negative three is straightforward.
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Itβs just zero as negative three is a constant.
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But what about the derivative of nine π₯ multiplied by nine π₯ square plus one to the power of negative one-half.
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Well, here, we have a product of differentiable functions.
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So, weβre going to need to apply the product rule.
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This tells us that for two differentiable functions π’ and π£, the derivative with respect to π₯ of their product π’π£ is equal to π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯.
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We multiply each function by the derivative of the other.
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So, we can define π’ to be the first factor, thatβs nine π₯, and π£ to be the second factor, thatβs nine π₯ squared plus one to the power of negative one-half.
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Using the power rule of differentiation, dπ’ by dπ₯ is equal to nine.
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And then, to find dπ£ by dπ₯, we need to use the chain rule extension to the power rule again.
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Which tells us that dπ£ by dπ₯ will be equal to negative a half multiplied by 18π₯ multiplied by nine π₯ squared plus one to the power of negative three over two.
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Weβve reduced the exponent by one.
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Using the product rule then, we have that the derivative with respect to π₯ of π’π£ is equal to π’ times dπ£ by dπ₯.
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Thatβs nine π₯ multiplied by negative a half 18π₯ multiplied by nine π₯ squared plus one to the negative three over two.
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Plus π£ times dπ’ by dπ₯, thatβs nine multiplied by nine π₯ squared plus one to the power of negative one-half.
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We can then take out a shared factor of nine multiplied by nine π₯ squared plus one to the power of negative three over two.
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What remains in the first term is negative nine π₯ squared.
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And what remains in the second is nine π₯ squared plus one to the power of one.
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Because, here, when we add the exponents of one and negative three over two, we get negative one-half.
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The largest set of parentheses simplifies nicely.
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We have negative nine π₯ squared plus nine π₯ square plus one, which all just simplifies to one.
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So, our derivative is simply nine multiplied by nine π₯ squared plus one to the power of negative three over two.
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And applying laws of exponents, we can rewrite this as nine over the square root of nine π₯ squared plus one cubed.
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So, returning to where we were writing down our expression for π double prime of π₯ then, we have that the second derivative is equal to zero plus nine over the square root of nine π₯ squared plus one cubed.
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And of course, we donβt need to include the zero.
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Remember that the purpose of this question is to determine where the function π of π₯ is concave up and concave down.
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And we recall that we said a functionβs concave up when its second derivative is greater than zero.
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So, letβs first consider where this is true.
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Well, this will be the case when nine over the square root of nine π₯ squared plus one cubed is greater than zero.
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Letβs consider this quotient.
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The numerator of this quotient is nine, which is a constant greater than zero.
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In the denominator, nine π₯ square plus one will give a value greater than or equal to one.
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And when we take its square root, by definition, weβre taking the positive square root, so we have a positive value.
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Weβre then cubing this, which will give another positive value.
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So, we see that, for all values of π₯, this quotient will be a positive value divided by a positive value, which will return a positive value.
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This means that the second derivative π double prime of π₯ is, in fact, always positive.
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And therefore, the function π of π₯ will be concave up throughout its domain, which in this case is the entire set of real numbers.
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Using interval notation then, we can conclude that the function π of π₯ will be concave up on the open interval from negative β to β, and there is no interval on which the function π of π₯ is concave down.