Example 3: f′(x) = 0 and f′(x) = DNE

Determine all relative extrema and the intervals where the function is increasing and decreasing.

$f\left(x\right)=x^{\frac{5}{3}}+10x^{\frac{2}{3}}$

Step 1: Find the 1 ^st Derivative, $f^{\prime }\left(x\right)$.

$f\left(x\right)=x^{\frac{5}{3}}+10x^{\frac{2}{3}}$

$f\prime \left(x\right)=\frac{5}{3}{x}^{\frac{2}{3}}+\frac{20}{3}{x}^{–\frac{1}{3}}$

Step 2: Find the critical values of the equation.

$f^{\prime }\left(x\right)=0$ OR $f^{\prime }\left(x\right)=\mathit{Does\; Not\; Exist}\left(\mathit{DNE}\right)$

First, notice that there is a negative exponent . You will want to rewrite that term to make the exponent positive .

After you perform the rewrite, you can see that there is actually a fraction in the problem. Anytime you have a fraction in your derivative, you must consider where you would get division by zero . Anytime that would occur you have a critical value from the $f^{\prime }\left(x\right)=\mathit{DNE}$ scenario.

To find ALL of your critical value you will need to solve two different algebra problems. The first is the one you will always do, which is to set the derivative equal to zero and solve for x . The second one is the one that most people forget, you must set just the bottom of our fraction equal to zero to see when the derivative would not exist , “break math”.

Algebra Moves $f^{\prime }\left(x\right)=0$:

i) First, we never want to solve for x on the bottom of a fraction. We will start by “clearing the fraction” and multiplying the entire equation by $x^{\frac{1}{3}}$. You would distribute it through each piece on the left-side of the equals. When you multiply x ’s you add the exponents .

ii) You now want to isolate the term with the x in it by subtracting $\frac{20}{3}$ from both sides of the equation.

iii) Multiply both sides of the equation by $\frac{3}{5}$to isolate the x on one side of the equation.

Algebra Moves $f^{\prime }\left(x\right)=\mathit{Does\; Not\; Exist}\left(\mathit{DNE}\right)$:

i) Grab only the piece on the bottom of the fraction. The piece that would cause division by zero to occur.

ii) Isolate the $x^{\frac{1}{3}}$ by dividing both sides by $3$.

iii) We now need to cancel out the $\frac{1}{3}$ power by making it a $1$. We want to get $x^1=x$. Since powers raised to powers is a multiply game, we need to raise both sides of the equation to the $3$ ^rd power.

$f\prime \left(x\right)=\frac{5}{3}{x}^{\frac{2}{3}}+\frac{20}{3}{x}^{–\frac{1}{3}}$

$f\prime \left(x\right)=\frac{5}{3}{x}^{\frac{2}{3}}+\frac{20}{3x^{\frac{1}{3}}}$

$f^{\prime }\left(x\right)=0$

$\frac{5}{3}{x}^{\frac{2}{3}}+\frac{20}{3x^{\frac{1}{3}}}=0$

$x^{\frac{1}{3}}\bullet \left(\frac{5}{3}{x}^{\frac{2}{3}}+\frac{20}{3x^{\frac{1}{3}}}\right)=\left(0\right)\bullet x^{\frac{1}{3}}$

$\frac{5}{3}x+\frac{20}{3}=0$

$\begin{array}{c}\frac{5}{3}x+\frac{20}{3}\\ \underline{–\frac{20}{3}}\end{array}\mathit{}\begin{array}{c}=0\\ \underline{–\frac{20}{3}}\end{array}$

$\frac{5}{3}x=–\frac{20}{3}$

$\frac{3}{5}\bullet \left(\frac{5}{3}x\right)=\left(–\frac{20}{3}\right)\bullet \frac{3}{5}$

$x=–4$

${\overline{)x=–4}}^{\mathit{Critical\; Value}}$

$f^{\prime }\left(x\right)=\mathit{Does\; Not\; Exist}\left(\mathit{DNE}\right)$

$f\prime \left(x\right)=\frac{5}{3}{x}^{\frac{2}{3}}+\frac{20}{\overline{)\mathrm{}3x^{\frac{1}{3}}}}$

$3x^{\frac{1}{3}}=0$

$\frac{3x^{\frac{1}{3}}}{3}=\frac{0}{3}$

$x^{\frac{1}{3}}=0$

${\left(x^{\frac{1}{3}}\right)}^3={\left(0\right)}^3$

$x=0$

${\overline{)x=0\mathrm{}}}^{\mathit{Critical\; Value}}$

Step 3: Do the actual 1 ^st Derivative Test (number line game).

i) Draw a number line.

ii) Mark all of your critical values on the number line.

iii) Choose test values on either side (left and right) of your critical values , and mark them on the number line.

iv) Do the actual 1 ^st Derivative Test. Plug your test values into your 1 ^st Derivative, $f\prime \left(x\right)$. Remember math people are not creative namers. It is called the 1 ^st Derivative Test it is because you are testing everything in the 1 ^st Derivative, $f\prime \left(x\right)$.

v) Mark the results either positive (increasing) or negative (decreasing) on your number line. Including drawing the directional arrow that goes with that behavior.

i)

ii)

iii)

iv) $f\prime \left(x\right)=\frac{5}{3}{x}^{\frac{2}{3}}+\frac{20}{3}{x}^{–\frac{1}{3}}$

$f\prime (\mathit{–}\mathit{8})=\frac{5}{3}{\left(\mathit{–}\mathit{8}\right)}^{\frac{2}{3}}+\frac{20}{3}{\left(\mathit{–}\mathit{8}\right)}^{–\frac{1}{3}}=\frac{10}{3}\left(\mathit{positive}\right)$

$f\prime (\mathit{–}\mathit{1})=\frac{5}{3}{\left(\mathit{–}\mathit{1}\right)}^{\frac{2}{3}}+\frac{20}{3}{\left(\mathit{–}\mathit{1}\right)}^{–\frac{1}{3}}=–5\left(\mathit{negative}\right)$

$f\prime \left(\mathit{1}\right)=\frac{5}{3}{\left(\mathit{1}\right)}^{\frac{2}{3}}+\frac{20}{3}{\left(\mathit{1}\right)}^{–\frac{1}{3}}=\frac{25}{3}\left(\mathit{positive}\right)$

v)

Step 4a Draw Conclusions : Relative Extrema

Reading off the relative extrema is really why I like drawing the directional arrows on the number line.

When you look for a relative maximum, we want to see the shape of a max .

When you look for a relative minimum, we want to see the shape of a min .

Looking at your two critical values you can see that we have a relative maximum occurring at x = -4 , a relative minimum occurring at x = 0 .

The final move is to now plug those x-values back into the original equation, $f\left(x\right)$. This will tell you what the actual max or min y-value is.

Step 4b Draw Conclusions: Intervals where the original equation, $f\left(x\right)$, is increasing and decreasing.

You will read this data directly from your 1 ^st Derivative Test number line . This is why I really like to draw the directional arrows on that number line.

Intervals on your 1 ^st Derivative Test number line where the derivative, $f\prime \left(x\right)$, value is positive, where you have an arrow pointing up are the increasing intervals .

Here you have a positive derivative starting at $–\mathrm{\infty }$and ending at our critical value of x = -4 . And starting again at your critical value of x = 0 and going on to $\infty$.

Intervals on your 1 ^st Derivative Test number line where the derivative, $f\prime \left(x\right)$, value is negative, where you have an arrow pointing down are the decreasing intervals .

Here you have a negative derivative starting at your critical value of x = -4 and ending at the critical value x = 0 .

Step 4a Draw Conclusions : Relative Extrema

$f\left(x\right)=x^{\frac{5}{3}}+10x^{\frac{2}{3}}$

Local Max: $f(–4)={(–4)}^{\frac{5}{3}}+10{(–4)}^{\frac{2}{3}}\approx 15.119$

Local Max: $f(–4)\approx 15.119$

Local Min: $f\left(0\right)={\left(0\right)}^{\frac{5}{3}}+10{\left(0\right)}^{\frac{2}{3}}=0$

Local Min: $f\left(0\right)=0$

Step 4b Draw Conclusions: Intervals where the original equation, $f\left(x\right)$, is increasing and decreasing.

The original function, $f\left(x\right)=x^{\frac{5}{3}}+10x^{\frac{2}{3}}$, is increasing on the x-interval $\left(–\infty ,–4\right)\cup \left(0,\infty \right)$.

The original function, $f\left(x\right)=x^{\frac{5}{3}}+10x^{\frac{2}{3}}$, is decreasing on the x-interval $\left(–4,0\right)$.

Final Result:

The function $f\left(x\right)=x^{\frac{5}{3}}+10x^{\frac{2}{3}}$ has a relative max of y = $15.11$ at x = -4 .

The function $f\left(x\right)=x^{\frac{5}{3}}+10x^{\frac{2}{3}}$ has a relative min of y = 0 at x = 0 .

The function $f\left(x\right)=x^{\frac{5}{3}}+10x^{\frac{2}{3}}$is increasing on the x-interval $\left(–\infty ,–4\right)\cup \left(0,\infty \right)$.

The function $f\left(x\right)=x^{\frac{5}{3}}+10x^{\frac{2}{3}}$, is decreasing on the x-interval $\left(–4,0\right)$.