Find the intercept, relative extrema, point of inflection, and asymptotes of the function(if they exist)of:
y= 3[(x-1)^(2/3)] - [(x-1)^2]
To find the intercepts of a function, we need to set y equal to zero and solve for x.
1. x-intercept:
Setting y = 0:
0 = 3[(x-1)^(2/3)] - [(x-1)^2]
This equation can be simplified to:
0 = (x-1)^2 * [3(x-1)^(2/3) - 1]
Now, we can set each factor equal to zero:
x-1 = 0 -> x = 1
3(x-1)^(2/3) - 1 = 0
3(x-1)^(2/3) = 1
(x-1)^(2/3) = 1/3
Taking both sides to the power of 3/2:
x-1 = (1/3)^(3/2)
x-1 = 1/√3
x = 1 + 1/√3
Therefore, the x-intercepts are x = 1 and x = 1 + 1/√3.
2. y-intercept:
To find the y-intercept, substitute x = 0 into the equation:
y = 3[(0-1)^(2/3)] - [(0-1)^2]
y = 3[-1^(2/3)] - 1
y = -3 - 1
y = -4
Therefore, the y-intercept is y = -4.
3. Relative Extrema:
To find the relative extrema, we need to find the critical points. These occur when the derivative of the function is equal to zero or undefined.
First, let's simplify the given function:
y = 3[(x-1)^(2/3)] - [(x-1)^2]
y = 3(x-1)^(2/3) - (x-1)^2
Now, let's find the derivative of the function:
y' = 2(x-1)^(-1/3) - 2(x-1)
To find the critical points, set y' = 0 and solve for x:
2(x-1)^(-1/3) - 2(x-1) = 0
Multiplying through by (x-1)^(3/2), we get:
2 - 2(x-1)^(3/2) = 0
Simplifying further:
2 = 2(x-1)^(3/2)
1 = (x-1)^(3/2)
Taking both sides to the power of 2/3:
1^(2/3) = (x-1)
1 = x-1
x = 2
Therefore, the critical point is x = 2. To determine whether it's a relative minimum or maximum, we can analyze the second derivative.
Taking the derivative of y' = 2(x-1)^(-1/3) - 2(x-1):
y'' = -2/(3(x-1)^(4/3)) - 2
Substituting x = 2 into the second derivative:
y''(2) = -2/(3(2-1)^(4/3)) - 2
y''(2) = -2/(3(1)^(4/3)) - 2
y''(2) = -2/(3) - 2
y''(2) = -2/3 - 2
y''(2) = -2/3 - 6/3
y''(2) = -8/3
Since y''(2) is negative, the function has a relative maximum at x = 2.
4. Point of Inflection:
To find the point(s) of inflection, we need to find where the concavity of the function changes. This happens when the second derivative is equal to zero or undefined.
Setting y'' = 0 and solving for x:
-2/(3(x-1)^(4/3)) - 2 = 0
Multiplying through by (x-1)^(4/3), we get:
-2 - 2(x-1)^(4/3) = 0
Simplifying further:
-2 = 2(x-1)^(4/3)
-1 = (x-1)^(4/3)
Taking both sides to the power of 3/4:
(-1)^(3/4) = (x-1)
Since (-1)^(3/4) is complex (i.e., imaginary), there are no real points of inflection for this function.
5. Asymptotes:
To find the asymptotes, we need to determine the behavior of the function as x approaches infinity and negative infinity.
As x approaches infinity, the terms involving (x-1) in the equation become negligible compared to the power of 2/3. Therefore, the function approaches positive infinity as x approaches infinity.
As x approaches negative infinity, the terms involving (x-1) in the equation become negligible compared to the power of 2/3. Therefore, the function approaches positive infinity as x approaches negative infinity.
Therefore, there are no horizontal asymptotes for this function.
In summary:
- x-intercepts: x = 1, x = 1 + 1/√3
- y-intercept: y = -4
- Relative Extrema: Relative maximum at x = 2
- Points of Inflection: None (no real points of inflection)
- Asymptotes: None (no horizontal asymptotes)