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:
#g'(-1) = 4(-1)^3 - 16(-1) = -4 + 16 = 12#
Since this is positive, the function is increasing on
Test point 2:
#g'(1) = 4(1)^3 - 16(1) = 4 - 16 = -12#
Since this is negative, the function is decreasing on
Test point 3:
#g'(3) = 4(3)^3 - 16(3) = 60#
Since this is positive, the function is increasing on
Test point 4:
#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:
#g''(-3) = 12(-3)^2 - 16 = 92#
Since this is positive, the function is concave up on
Test point 2:
#g''(0) = 12(0)^2 - 16 = -16#
Since this is negative, the function is concave down on
Test point 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!