Question

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

Answer 1

Start by finding the first derivative.

`g'(x) = 4x^3 - 16x`

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

`0 = 4x^3 - 16x`

`0 = 4x(x^2 - 4)`

`0 = 4x(x + 2)(x - 2)`

`x = 0, 2, and -2`

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

`g''(x) = 12x^2 - 16`

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

`0 = 12x^2 - 16`

`0 = 4(3x^2 - 4)`

`x = +- 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'(a) > 0`, then `g(x)` is increasing at that point, and if `g'(a) < 0`, then `g(x)` is decreasing at that point.

Select test points between the critical points.

Test point 1: `x = -1`

`g'(-1) = 4(-1)^3 - 16(-1) = -4 + 16 = 12`

Since this is positive, the function is increasing on `(-2, 0)`.

Test point 2: `x = 1`

`g'(1) = 4(1)^3 - 16(1) = 4 - 16 = -12`

Since this is negative, the function is decreasing on `(0, 2)`

Test point 3: `x = 3`

`g'(3) = 4(3)^3 - 16(3) = 60`

Since this is positive, the function is increasing on `(2, oo)`.

Test point 4: `x = -3`

`g'(-3) = 4(-3)^3 - 16(-3) = -60`

Since this is negative, the function is decreasing on `(-oo, -2)`.

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

We will once again select test points.

Test point 1: `x = -3`

`g''(-3) = 12(-3)^2 - 16 = 92`

Since this is positive, the function is concave up on `(-oo, -2/sqrt(3))`.

Test point 2: `x = 0`

`g''(0) = 12(0)^2 - 16 = -16`

Since this is negative, the function is concave down on `(-2/sqrt(3), 2/sqrt(3))`.

Test point 3: `x = 3`

`g''(3) = 12(3)^3 - 16 = 92`

Since this is positive, the function is concave up on `(2/sqrt(3), oo)`.

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 - 8x^2 - 10`

We let `u = x^2`.

`0 = u^2 - 8u - 10`

`u = (-(-8) +- sqrt((-8)^2 - 4 xx 1 xx -10))/(2 xx 1)`

`u = (8 +- 2sqrt(26))/2`

`u = 4 +- sqrt(26)`

We now revert to `x`.

`x^2 = 4 +- sqrt(26)`

`x = +- sqrt(4 +- sqrt(26))`

But since the value under the `√` must always be positive, the `+-` sign should become a negative a `+`.

`x = +- sqrt(4 + sqrt(26))`

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

`g(0) = 0^4 - 8(0)^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!