Section 3.3 : Differentiation Formulas
14. Determine where, if anywhere, the function \(y = 2{z^4} - {z^3} - 3{z^2}\) is not changing.
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Recall that one of the interpretations of the derivative is that it gives the rate of change of the function. So, the function won’t be changing if its rate of change is zero and so all we need to do is find the derivative and set it equal to zero to determine where the rate of change is zero and hence the function will not be changing.
First the derivative, and we’ll do a little factoring while we are at it.
\[\frac{{dy}}{{dz}} = 8{z^3} - 3{z^2} - 6z = z\left( {8{z^2} - 3z - 6} \right)\] Show Step 2Now all that we need to do is set this equation to zero and solve.
\[\begin{align*}\frac{{dy}}{{dz}} & = 0\\ & z\left( {8{z^2} - 3z - 6} \right) = 0\hspace{0.25in}\,\,\,\,\, \to \hspace{0.25in}\hspace{0.25in}z = 0,\,\,\,\,\,\,\,\,\,8{z^2} - 3z - 6 = 0\end{align*}\]We can easily see from this that the derivative will be zero at \(z = 0\), however, because the quadratic doesn’t factor we’ll need to use the quadratic formula to determine where, if anywhere, that will be zero.
\[z = \frac{{3 \pm \sqrt {{{\left( { - 3} \right)}^2} - 4\left( 8 \right)\left( { - 6} \right)} }}{{2\left( 8 \right)}} = \frac{{3 \pm \sqrt {201} }}{{16}}\]The function therefore not be changing at,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{z = 0 \hspace{0.25in} z = \frac{{3 + \sqrt {201} }}{{16}} = 1.07359 \hspace{0.25in} z = \frac{{3 - \sqrt {201} }}{{16}} = - 0.69859}}\]