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### Section 4.2 : Critical Points

9. Determine the critical points of $$r\left( y \right) = \sqrt{{{y^2} - 6y}}$$.

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We’ll need the first derivative to get the answer to this problem so let’s get that.

$r'\left( y \right) = {\textstyle{1 \over 5}}\left( {2y - 6} \right){\left( {{y^2} - 6y} \right)^{ - \,\,\frac{4}{5}}} = \frac{{2y - 6}}{{5{{\left( {{y^2} - 6y} \right)}^{\frac{4}{5}}}}}$

We took the term with the negative exponent to the denominator for the discussion in the next step. While it doesn’t really need to be done this will make sure that there are no inadvertent mistakes down the road.

Show Step 2

Recall that critical points are simply where the derivative is zero and/or doesn’t exist. In this case the derivative is a rational expression. Therefore, we know that the derivative will be zero if the numerator is zero (and the denominator is also not zero for the same values of course). We also know that the derivative won’t exist if we get division by zero.

So, all we need to do is set the numerator and denominator equal to zero and solve. Note that the exponent on the whole denominator will not affect where it is zero and so can be ignored. This means we need to solve the following two equations.

$2y - 6 = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}y = 3$ ${y^2} - 6y = y\left( {y - 6} \right) = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}y = 0,6$ Show Step 3

Note as well that the reason for moving the term to the denominator as we did in the first step is to make it clear that the last two critical points are critical points because the derivative does not exist at those points and not because the derivative is zero at those points. Also note that they are critical points because the function does exist at these points.

Therefore, along with the first critical point (where the derivative is zero), we get the following critical points for this function.

$\require{bbox} \bbox[2pt,border:1px solid black]{{y = 0,\,\,3,\,\,6}}$