Consider the function h(x) = a(-2x+1)^5-b, where a doesn't equal 0 and b doesn't equal 0 are constants.
A. Find h'(x) and h"(x)
B. show that h is monotonic ( that is, that either h always increases or remains constant or h always decreases or remains constant)
C. Show that x-coordinate(s) of the location(s) of the critical points are independent of a and b.
h' = 5a(-2x+1)^4 (-2) = -10a(-2x+1)^4
h'' = 80a(-2x+1)^3
Since h' >=0 for all x, it is monotonic
h'=0 ==> (-2x+1)^4 = 0, independent of a,b
A. To find the derivatives of the function h(x), we need to apply the chain rule.
1. h'(x):
The power rule states that the derivative of x^n with respect to x is n*x^(n-1).
Using the chain rule, we have:
h'(x) = a * 5(-2x + 1)^(5-1) * (-2)
= -10a(-2x + 1)^4
2. h"(x):
We need to find the derivative of h'(x):
h"(x) = -10a * 4(-2x + 1)^(4-1) * (-2)
= 40a(-2x + 1)^3
B. To determine whether the function h(x) is monotonic, we need to analyze the sign of its derivative.
1. h'(x):
Since a and b are constants and we don't have information about their values, we cannot make conclusions about the sign of h'(x) directly.
C. To show that the x-coordinate(s) of the critical points are independent of a and b, we need to find the values of x for which h'(x) equals zero (critical points).
1. Set h'(x) to zero and solve for x:
-10a(-2x + 1)^4 = 0
We can see that the only factor that affects the values of x is (-2x + 1). This means that no matter what values a and b have, the x-coordinates of the critical points will be the same.
To answer these questions, we need to find the derivatives of the function h(x) with respect to x.
A. Find h'(x) and h''(x)
To find the derivative of h(x), we'll use the chain rule. Let's step-by-step find the derivatives:
Step 1: Differentiate the outer function.
h(x) = a(-2x+1)^5-b
The outer function is (-2x+1)^5. The derivative of this function is:
d/dx [(-2x+1)^5] = 5(-2x+1)^4
Step 2: Multiply by the derivative of the inner function.
The inner function is -2x+1. The derivative of this function is:
d/dx [-2x+1] = -2
Step 3: Apply the chain rule.
By the chain rule, we multiply the derivative of the outer function by the derivative of the inner function.
h'(x) = (5(-2x+1)^4)(-2)
Similarly, to find the second derivative h''(x), we differentiate h'(x) with respect to x:
h''(x) = d/dx[h'(x)] = d/dx[(5(-2x+1)^4)(-2)]
B. Show that h is monotonic
To show that h is monotonic, we need to determine whether its derivative h'(x) is always positive or always negative or whether it remains constant on its domain.
From the derivative h'(x), we can see that it doesn't depend on the values of a and b, so we can ignore them for this analysis.
To determine the sign of h'(x), we need to consider when h'(x) > 0 and when h'(x) < 0.
If h'(x) > 0, it means that the function h(x) is increasing.
If h'(x) < 0, it means that the function h(x) is decreasing.
If h'(x) = 0, it means that the function h(x) is constant.
From the expression for h'(x), we can see that it is always negative (-2 times a positive number raised to an even power). Therefore, h(x) is always decreasing, or in other words, it is monotonic.
C. Show that the x-coordinate(s) of the critical points are independent of a and b.
To find the critical points, we need to find the values of x for which h'(x) = 0.
Setting h'(x) = 0 and solving for x:
0 = 5(-2x+1)^4(-2)
Since the derivative h'(x) is always negative, there are no critical points—where the derivative is zero—for this function. Therefore, the x-coordinate(s) of the critical points are independent of a and b, as there are none.