We know that the derivative of a function gives us the rate of change of the function and so we’ll first need the derivative of this function. We computed this derivative in Problem 10 from the previous section and so we won’t show the work here. If you need the practice you should go back and redo that problem before proceeding.

So, from our previous work we know that the derivative is,

Now all that we need to do is to compute : \(Z'\left( 5 \right) = \frac{3}{{2\sqrt {11} }}\). This is positive and so we know that the function must be increasing at \(t = 5\).

Again, all we need to do is compute a derivative and since we’ve got the derivative written down in the first part there’s no reason to redo that here.

The evaluation is : \(Z'\left( {10} \right) = \frac{3}{{2\sqrt {26} }}\). This is positive and so we know that the function must be increasing at \(t = 10\).

Again, all we need to do is compute a derivative and since we’ve got the derivative written down in the first part there’s no reason to redo that here.

The evaluation is : \(Z'\left( {300} \right) = \frac{3}{{2\sqrt {896} }}\). This is positive and so we know that the function must be increasing at \(t = 300\).

Final Note

As a final note to all the parts of this problem let’s notice that we did not really need to do any evaluations. Because we know that square roots will always be positive it is clear that the derivative will always be positive regardless of the value of \(t\) we plug in.