Section 4.2 : Critical Points
10. Determine the critical points of \(h\left( t \right) = 15 - \left( {3 - t} \right){\left[ {{t^2} - 8t + 7} \right]^{\frac{1}{3}}}\).
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Start SolutionWe’ll need the first derivative to get the answer to this problem so let’s get that.
\[\begin{align*}h'\left( t \right) & = {\left[ {{t^2} - 8t + 7} \right]^{\frac{1}{3}}} - \left( {3 - t} \right)\left( {{\textstyle{1 \over 3}}} \right)\left( {2t - 8} \right){\left[ {{t^2} - 8t + 7} \right]^{ - \,\,\frac{2}{3}}} = {\left[ {{t^2} - 8t + 7} \right]^{\frac{1}{3}}} - \frac{{\left( {3 - t} \right)\left( {2t - 8} \right)}}{{3{{\left[ {{t^2} - 8t + 7} \right]}^{\,\frac{2}{3}}}}}\\ & = \frac{{3\left( {{t^2} - 8t + 7} \right) - \left( {3 - t} \right)\left( {2t - 8} \right)}}{{3{{\left( {{t^2} - 8t + 7} \right)}^{\,\frac{2}{3}}}}} = \frac{{5{t^2} - 38t + 45}}{{3{{\left( {{t^2} - 8t + 7} \right)}^{\,\frac{2}{3}}}}}\end{align*}\]After differentiating we moved the term with the negative exponent to the denominator and then combined everything into a single term. This will help with the next step considerably.
Show Step 2Recall that critical points are simply where the derivative is zero and/or doesn’t exist.
Because we moved the term with the negative exponent to the denominator and then combined everything into a single term we now have written the derivative as 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.
\[5{t^2} - 38t + 45 = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}t = \frac{{38 \pm \sqrt {544} }}{{10}} = \frac{{19 \pm 2\sqrt {34} }}{5}\] \[{t^2} - 8t + 7 = \left( {t - 7} \right)\left( {t - 1} \right) = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}t = 1,\,\,7\] Show Step 3Note that because we combined all the terms in the derivative into a single term it was much easier to determine the critical points for this function. If we had not combined the terms the solving work would have been more complicated, although not impossible.
Doing this also makes 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 two critical points (where the derivative is zero), we get the following critical points for this function.
\[\require{bbox} \bbox[2pt,border:1px solid black]{{t = 1,\,\,7,\,\,\frac{{19 \pm 2\sqrt {34} }}{5}}}\]