Finding the Shape of a Graph Using the Min and Max

Question

QUESTION: Let $f(x)=2x^{2}-2x^{4}$. Find the open intervals on which $f$ is increasing (decreasing). Then determine the x-coordinates of all relative maxima (minima).

1. $f$ is increasing on the intervals: __________
2. $f$ is decreasing on the intervals: __________
3. The relative maxima of $f$ occur at x = __________
4. The relative minima of $f$ occur at x = __________

I have tried to solve the problem via the method below:

$f'(x)=4x-8x^3$

$f'(x)=4(x-x^3)$

$f'(x)=4(x-x^3)=0$ when $x=1,0,-1$

$f(-1) = 0 \hspace{10mm} f(0) = 0 \hspace{10mm} f(1) = 0$

$f''(x) = 4-24x^2$

$f''(x) = \frac{0\pm\sqrt{0-4(-24)(4)}}{-48}$

$f''(x) = \pm\frac{\sqrt{6}}{6}$

Since $f(-1) = 0 , \hspace{3mm} f(0) = 0 ,\hspace{3mm}f(1) = 0$ I concluded that there was no relative maxima or minima as the values for the critical points, when plugged into the initial function $f(x)$, were the same, 0.

To find the interval where the function was increasing I used the points of inflection ($\pm\frac{\sqrt{6}}{6}$). I plugged in -2, .1 and 2 to the original function. I then concluded that $f(-2)$ would yield a negative number, $f(.1)$ would yield a positive number, and $f(2)$ would yield a negative number.

MY ANSWER

1. $f$ is increasing on the intervals: $(-\frac{\sqrt{6}}{6}, \frac{\sqrt{6}}{6})$

2. $f$ is decreasing on the intervals: $(-\infty,-\frac{\sqrt{6}}{6})\cup(\frac{\sqrt{6}}{6},\infty)$

3. The relative maxima of $f$ occur at x = none

4. The relative minima of $f$ occur at x = none

CORRECT ANSWERS

1. $f$ is increasing on the intervals: $\left( - \infty , - \left( \sqrt { \frac { 2 } { 2 \cdot 2 } } \right) \right) , \left( 0 , \sqrt { \frac { 2 } { 2 \cdot 2 } } \right)$

2. $f$ is decreasing on the intervals: $\left( - \left( \sqrt { \frac { 2 } { 2 \cdot 2 } } \right) , 0 \right) , \left( \sqrt { \frac { 2 } { 2 \cdot 2 } } , \infty \right)$

3. The relative maxima of $f$ occur at x = −0.707107 , 0.707107

4. The relative minima of $f$ occur at x = 0

I would like to understand why my answers are incorrect. If someone could take a look at my steps and explain to me where I went wrong it would be greatly appreciated!

Answer 1

So, let's take a look at what's going on.

You found $f'(x)$ correctly:

$f'(x) = 4x - 8x^3$

But then if we factor out a $4x$ we obtain $f'(x) = 4x(1-2x^2)$. Clearly, $f'(x) = 0$ when $x = 0$, and solving $1-2x^2 =0$ allows us to find that $x = \frac{\sqrt2}{2}$ and $x = -\frac{\sqrt2}{2}$ are also critical points. Plugging these values of $x$ into $f(x)$, we obtain:

$f(0) = 0,\quad f(\frac{\sqrt2}{2}) = \frac{1}{2}, \quad f(-\frac{\sqrt2}{2}) = \frac{1}{2}$

In order to find the relative maxima and relative minima, we must plug in values of x into $f'(x)$ that are between $-\frac{\sqrt2}{2}$ and $0$, between $0$ and $\frac{\sqrt2}{2}$, before $-\frac{\sqrt2}{2}$, and after $\frac{\sqrt2}{2}$. We want to know what the sign of our derivative is during these intervals. Do you think you can take it from here?