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### Section 4.5 : The Shape of a Graph, Part I

15. Given that \(f\left( x \right)\) is an increasing function and define \(h\left( x \right) = {\left[ {f\left( x \right)} \right]^2}\). Will \(h\left( x \right)\) be an increasing function? If yes, prove that \(h\left( x \right)\) is an increasing function. If not, can you determine any other conditions needed on the function \(f\left( x \right)\) that will guarantee that \(h\left( x \right)\) will also increase?

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We know that the derivative can be used to tell us if a function is increasing so let’s find the derivative of \(h\left( x \right)\). Do not get excited about the fact that we don’t know what \(f\left( x \right)\) is. We can symbolically take the derivative with a quick application of the chain rule.

Here is the derivative of \(h\left( x \right)\).

\[h'\left( x \right) = 2{\left[ {f\left( x \right)} \right]^1}f'\left( x \right) = 2f\left( x \right)f'\left( x \right)\]We know that a function will be increasing if its derivative is positive. So, the question we need to answer is can we guarantee that \(h'\left( x \right) > 0\) if we only take into account the assumption that \(f\left( x \right)\) is an increasing function.

From our assumption that \(f\left( x \right)\) is an increasing function and so we know that \(f'\left( x \right) > 0\).

Now, let’s see what \(h'\left( x \right)\) tells us. We can see that \(h'\left( x \right)\) is a product of a number and two functions. The “2” is positive and so the sign of the derivative will come from the sign of the product of \(f\left( x \right)\) and \(f'\left( x \right)\).

Okay, from our assumption we know that \(f'\left( x \right)\) is positive. However, the product isn’t guaranteed to be positive.

For example, consider \(f\left( x \right) = x\). Clearly, \(f'\left( x \right) = 1 > 0\), and so this is an increasing function. However, \(f\left( x \right)f'\left( x \right) = \left( x \right)\left( 1 \right) = x\). Therefore, we can see that the product will not always be positive. This shouldn’t be too surprising given that,

\[h\left( x \right) = {\left[ {f\left( x \right)} \right]^2} = {\left[ x \right]^2} = {x^2}\]In this case we can clearly see that \(h\left( x \right)\) will not always be an increasing function.

On the other had if we let \(f\left( x \right) = {{\bf{e}}^x}\) we can see that \(f'\left( x \right) = {{\bf{e}}^x} > 0\) and we can also see that \(f\left( x \right)f'\left( x \right) = \left( {{{\bf{e}}^x}} \right)\left( {{{\bf{e}}^x}} \right) = {{\bf{e}}^{2x}} > 0\).

So, from these two examples we can see that we can find increasing functions, \(f\left( x \right)\), for which \({\left[ {f\left( x \right)} \right]^2}\) may or may not always be increasing.

So, just what was the difference between the two examples above?

The problem with the first example, \(f\left( x \right) = x\) was that it wasn’t always positive and so the product of \(f\left( x \right)\) and \(f'\left( x \right)\) would not always be positive as we needed it to be.

This was not a problem with the second example however, \(f\left( x \right) = {{\bf{e}}^x}\). In this case the function is always positive and so the product of the function and its derivative will also be positive.

That is also the added condition that we need to guarantee that \(h\left( x \right)\) will be positive.

If we start with the assumption that \(f\left( x \right)\) in an increasing function we need to further assume that \(f\left( x \right)\) is a positive function in order to guarantee that \(h\left( x \right) = {\left[ {f\left( x \right)} \right]^2}\) will be an increasing function.