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

##### 1 Answer

Start by finding the first derivative.

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

Now find the critical numbers, which will occur when the derivative

#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 = 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

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

**Test point 2: #x = 1#**

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

Since this is negative, the function is decreasing on

**Test point 3: #x = 3#**

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

Since this is positive, the function is increasing on

**Test point 4: #x = -3#**

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

Since this is negative, the function is decreasing on

Now we go back to the second derivative to check the concave up/concave down intervals. If

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

**Test point 2: #x = 0#**

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

Since this is negative, the function is concave down on

**Test point 3: #x = 3#**

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

Since this is positive, the function is concave up on

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

#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^2 = 4 +- sqrt(26)#

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

But since the value under the

#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!