function f(x)=x3x2−1 defined on the interval [−16,16]
A. Find the vertical asymptotes of function f(x)
B. Find where it is concave up
C. Find the inflection points
To find the vertical asymptotes of the function f(x) = x^3x^2 - 1, we need to analyze the behavior of the function as x approaches certain values.
A. To find the vertical asymptotes:
1. Determine the values of x that would make the denominator of the rational function equal to zero. In this case, the denominator is x^2.
x^2 = 0
x = 0
So, x = 0 is a vertical asymptote of the function f(x).
B. To find where the function is concave up:
1. Compute the second derivative of the function f(x).
f''(x) = (d^2/dx^2)[x^3x^2 - 1]
f''(x) = (6x^5 + 6x^3)
2. Set the second derivative equal to zero and solve for x to find the critical points.
6x^5 + 6x^3 = 0
6x^3(x^2 + 1) = 0
Set each term equal to zero:
6x^3 = 0 --> x = 0 (This is the same value as the vertical asymptote)
x^2 + 1 = 0 --> x^2 = -1
Since the square of a real number cannot be negative, there are no real solutions for x^2 + 1 = 0.
Therefore, we have only one critical point, x = 0.
C. To find the inflection points:
1. Compute the second derivative of the function f(x). (We found it earlier)
f''(x) = 6x^5 + 6x^3
2. Set the second derivative equal to zero and solve for x to find the critical points.
6x^5 + 6x^3 = 0
6x^3(x^2 + 1) = 0
Set each term equal to zero:
6x^3 = 0 --> x = 0 (This is the same value as the vertical asymptote)
x^2 + 1 = 0 --> x^2 = -1
Again, there are no real solutions for x^2 + 1 = 0.
Since we only have one critical point, x = 0, there are no inflection points in the given interval [-16, 16].