How do you find the critical points to sketch the graph g(x)=x^4-8x^2-10?

Apr 1, 2017

Start by finding the first derivative.

$g ' \left(x\right) = 4 {x}^{3} - 16 x$

Now find the critical numbers, which will occur when the derivative $=$ $0$.

$0 = 4 {x}^{3} - 16 x$

$0 = 4 x \left({x}^{2} - 4\right)$

$0 = 4 x \left(x + 2\right) \left(x - 2\right)$

$x = 0 , 2 , \mathmr{and} - 2$

We now determine the second derivative to find the points of inflection.

$g ' ' \left(x\right) = 12 {x}^{2} - 16$

Once again, we set the derivative to $0$ and solve (except this time it'll be the second derivative).

$0 = 12 {x}^{2} - 16$

$0 = 4 \left(3 {x}^{2} - 4\right)$

$x = \pm \frac{2}{\sqrt{3}}$

Now let's revert our attention back to the 1st derivative. We must determine the intervals of increase decrease. Quite simply, if $g ' \left(a\right) > 0$, then $g \left(x\right)$ is increasing at that point, and if $g ' \left(a\right) < 0$, then $g \left(x\right)$ is decreasing at that point.

Select test points between the critical points.

Test point 1: $x = - 1$

$g ' \left(- 1\right) = 4 {\left(- 1\right)}^{3} - 16 \left(- 1\right) = - 4 + 16 = 12$

Since this is positive, the function is increasing on $\left(- 2 , 0\right)$.

Test point 2: $x = 1$

$g ' \left(1\right) = 4 {\left(1\right)}^{3} - 16 \left(1\right) = 4 - 16 = - 12$

Since this is negative, the function is decreasing on $\left(0 , 2\right)$

Test point 3: $x = 3$

$g ' \left(3\right) = 4 {\left(3\right)}^{3} - 16 \left(3\right) = 60$

Since this is positive, the function is increasing on $\left(2 , \infty\right)$.

Test point 4: $x = - 3$

$g ' \left(- 3\right) = 4 {\left(- 3\right)}^{3} - 16 \left(- 3\right) = - 60$

Since this is negative, the function is decreasing on $\left(- \infty , - 2\right)$.

Now we go back to the second derivative to check the concave up/concave down intervals. If $g ' ' \left(a\right) > 0$, then $g \left(x\right)$ is concave up at that point, and if $g ' ' \left(a\right) < 0$, then $g \left(x\right)$ is concave down at that point.

We will once again select test points.

Test point 1: $x = - 3$

$g ' ' \left(- 3\right) = 12 {\left(- 3\right)}^{2} - 16 = 92$

Since this is positive, the function is concave up on $\left(- \infty , - \frac{2}{\sqrt{3}}\right)$.

Test point 2: $x = 0$

$g ' ' \left(0\right) = 12 {\left(0\right)}^{2} - 16 = - 16$

Since this is negative, the function is concave down on $\left(- \frac{2}{\sqrt{3}} , \frac{2}{\sqrt{3}}\right)$.

Test point 3: $x = 3$

$g ' ' \left(3\right) = 12 {\left(3\right)}^{3} - 16 = 92$

Since this is positive, the function is concave up on $\left(\frac{2}{\sqrt{3}} , \infty\right)$.

It's true that you could just have found the next intervals of concavity and increasing/decreasing after the first by following the pattern of positive-negative-positive/negative-positive-negative, depending on the first interval, but I wanted to show you this method to make it as clear as possible.

The last thing I would like to discuss before graphing is intercepts. First, for the x-intercepts.

$0 = {x}^{4} - 8 {x}^{2} - 10$

We let $u = {x}^{2}$.

$0 = {u}^{2} - 8 u - 10$

$u = \frac{- \left(- 8\right) \pm \sqrt{{\left(- 8\right)}^{2} - 4 \times 1 \times - 10}}{2 \times 1}$

$u = \frac{8 \pm 2 \sqrt{26}}{2}$

$u = 4 \pm \sqrt{26}$

We now revert to $x$.

${x}^{2} = 4 \pm \sqrt{26}$

$x = \pm \sqrt{4 \pm \sqrt{26}}$

But since the value under the √ must always be positive, the $\pm$ sign should become a negative a $+$.

$x = \pm \sqrt{4 + \sqrt{26}}$

Finally, as for the y-intercept, we have:

$g \left(0\right) = {0}^{4} - 8 {\left(0\right)}^{2} - 10 = - 10$

We now trace the following graph putting all of the previous elements together.

graph{x^4 - 8x^2 - 10 [-58.5, 58.5, -29.27, 29.3]}

Hopefully this helps!