Find the relative maximum and relative minimum of the function $f(x) = 49x + \frac{4}{x}$

Question

Find the relative maximum and relative minimum of the function $f(x) = 49x + \frac{4}{x}=49x+4x^{-1}$

Solution:

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Step 1: Find the values of $x$ where $f'(x)=0$ and $f'(x)$ DNE.

$f'(x)=49-4x^{-2}=49-\frac{4}{x^2}$

We can see that $f'(x)$ DNE when $x=0$. Now lets find the values of $x$ that make $f'(x)=0$.

$f'(x)=49-4x^{-2}=49-\frac{4}{x^2}=0$

$\rightarrow x^2=\frac{4}{49}$

$\rightarrow x= \pm \sqrt{\frac{4}{49}}$

$\rightarrow x= \pm \frac{2}{7}$

Thus our critical points are $x=0, x= \frac{2}{7}, x= -\frac{2}{7}$

Step 2: Make a number line and plot the sign of $f'(x)$ to find where $f(x)$ is increasing and decreasing.

$f'(-3)= 49-\frac{4}{9}>0$

$f'(\frac{-1}{7})=49-\frac{4}{(\frac{-1}{7})^2}=49-4(49)=-3(49)<0$

$f'(\frac{1}{7})=49-\frac{4}{(\frac{1}{7})^2}=49-4(49)=-3(49)<0$

$f'(3)=49-\frac{4}{3^2} >0$

Thus $f(x)$ is increasing as $x$ increases towards $\frac{-2}{7}$ and then $f(x)$ is decreasing as $x$ increases towards $0$. Therefore $x=\frac{-2}{7}$ corresponds to a relative maximum.

Thus $f(x)$ is decreasing $x$ increases towards $\frac{2}{7}$ and then $f(x)$ is increases as $x$ increases towards $\infty$. Therefore $x=\frac{2}{7}$ corresponds to a relative minimum.

Step 3: find the corresponding $y$ values.

$f(\frac{2}{7})=28$

$f(\frac{-2}{7})=-28$

Therefore $(\frac{2}{7},28)$ is a relative minimum and $(\frac{-2}{7},-28)$ is a relative maximum.

This may be weird that our relative maximum is smaller than our relative minimum, but since these are "relative" maximum and minimum, they are only being compared to points on the graph very near to it, so this is okay.

Answer 1

First note that $f(-x)=-f(x)$, so the problem is essentially reduced to $x\ge0$.

That can be solved without calculus: by the AM-GM inequality, $49x+\dfrac4x\ge2\sqrt{49\times4}=28,$

and equality holds when $x=\dfrac27$.

Answer 2

Rewrite as \begin{eqnarray*} 49x+\frac{4}{x} = \left( 7 \sqrt{x} - \frac{2}{\sqrt{x}} \right)^2 +28. \end{eqnarray*} From this it easy to see that a minimum of $28$ occurs at $x=2/7$.

Also \begin{eqnarray*} 49x+\frac{4}{x} = \left( 7 \sqrt{x} + \frac{2}{\sqrt{x}} \right)^2 -28. \end{eqnarray*} similarly a maximum of $-28$ occurs at $x=-2/7$.

$0$ is not a turning point.