What is the inverse of each of the functions defined by the following equations, if the inverse
exists. If the inverse does not exist for the largest possible domain, limit the domain so that the inverse
will exist. In each case, give the domain and range of both the function and its inverse.
a. f(x) = x^2n + x^n + 1
b. g(x) = (x^2-4)^(1/2)
c. h(x) = (x)/(x^3)
d. S(x) = 1/2 (e^x - e^-x)
a. To find the inverse of the function f(x) = x^2n + x^n + 1, we can rearrange the equation and solve for x.
Step 1: Start with the equation f(x) = x^2n + x^n + 1.
Step 2: Replace f(x) with y, so the equation becomes y = x^2n + x^n + 1.
Step 3: Swap x and y in the equation, so we have x = y^2n + y^n + 1.
Step 4: Now, solve for y by rearranging the equation:
x - 1 = y^2n + y^n.
Step 5: We can see that this equation is not easily solvable for y. Therefore, the inverse of f(x) = x^2n + x^n + 1 does not exist for the largest possible domain.
Domain of f(x): All real numbers.
Range of f(x): All real numbers greater than or equal to 1.
Domain of f^(-1)(x): Not defined (since the inverse does not exist).
Range of f^(-1)(x): Not defined.
b. To find the inverse of the function g(x) = (x^2 - 4)^(1/2), we can follow a similar process.
Step 1: Start with the equation g(x) = (x^2 - 4)^(1/2).
Step 2: Replace g(x) with y, so the equation becomes y = (x^2 - 4)^(1/2).
Step 3: Swap x and y in the equation, so we have x = (y^2 - 4)^(1/2).
Step 4: Now, solve for y by rearranging the equation:
y = (x^2 - 4)^(1/2).
Step 5: Here, the range of the function g(x) is non-negative numbers (including zero), since taking the square root results in positive values. Therefore, the domain of the inverse function f^(-1)(x) should be non-negative numbers.
Domain of g(x): All real numbers.
Range of g(x): Non-negative numbers (including zero).
Domain of g^(-1)(x): Non-negative numbers (including zero).
Range of g^(-1)(x): All real numbers.
c. To find the inverse of the function h(x) = x/(x^3), we can proceed as follows:
Step 1: Start with the equation h(x) = x/(x^3).
Step 2: Replace h(x) with y, so the equation becomes y = x/(x^3).
Step 3: Swap x and y in the equation, so we have x = y/(y^3).
Step 4: Now, solve for y by rearranging the equation:
y = 1/x^2.
Step 5: Here, the domain of the function h(x) does not include x = 0, as it would cause division by zero. Therefore, the domain of the inverse function h^(-1)(x) should exclude x = 0.
Domain of h(x): All real numbers except x = 0.
Range of h(x): All real numbers.
Domain of h^(-1)(x): All real numbers except x = 0.
Range of h^(-1)(x): All real numbers.
d. To find the inverse of the function S(x) = 1/2 (e^x - e^(-x)), we can follow a similar process.
Step 1: Start with the equation S(x) = 1/2 (e^x - e^(-x)).
Step 2: Replace S(x) with y, so the equation becomes y = 1/2 (e^x - e^(-x)).
Step 3: Swap x and y in the equation, so we have x = 1/2 (e^y - e^(-y)).
Step 4: Now, solve for y by rearranging the equation:
2x = e^y - e^(-y).
Step 5: Unfortunately, there is no simple algebraic way to solve this equation for y. Therefore, the inverse of S(x) = 1/2 (e^x - e^(-x)) does not have an easily expressible form.
Domain of S(x): All real numbers.
Range of S(x): All real numbers.
Domain of S^(-1)(x): Not defined (since the inverse does not exist).
Range of S^(-1)(x): Not defined.
To find the inverse of a function, you need to switch the roles of x and y and solve the resulting equation for y.
a. For the function f(x) = x^2n + x^n + 1, we switch x and y, giving us x = y^2n + y^n + 1. Let's solve for y:
x = y^2n + y^n + 1
x - 1 = y^2n + y^n
Since this equation involves both y^n and y^2n terms, the inverse of this function will be more complicated. The inverse exists for a limited domain because if the exponent n is not an integer, then there may be values of x for which the equation has no real solutions for y. To ensure the inverse exists, we need to restrict the domain of f(x) to values of x that make y^n and y^2n real. The domain for the function f(x) is restricted to values where y^n and y^2n are real numbers.
The range of f(x) will depend on the value of n. For even values of n, f(x) will always be positive and have a minimum value of 1. For odd values of n, the range of f(x) will be all real numbers.
The domain and range of the inverse function will depend on the domain and range of f(x) after considering the restrictions for y^n and y^2n to be real numbers.
b. For the function g(x) = (x^2-4)^(1/2), we switch x and y, giving us x = (y^2-4)^(1/2). Let's solve for y:
x = (y^2-4)^(1/2)
x^2 = y^2 - 4
y^2 = x^2 + 4
Since the square root of a number always gives a positive result, the inverse exists for the entire domain of g(x), which is all real numbers.
The range of g(x) is non-negative numbers because the square root of any negative number is not defined. The range of g(x) is [0, ∞).
The domain and range of the inverse function will be the same as the domain and range of g(x).
c. For the function h(x) = x/(x^3), we switch x and y, giving us x = y/(y^3). Let's solve for y:
x = y/(y^3)
xy^3 = y
To solve for y, we can divide both sides of the equation by x, giving us:
y^3 = 1/x
y = (1/x)^(1/3)
y = x^(-1/3)
The inverse exists for the entire domain of h(x) because the equation has a real solution for y for any value of x.
The range of h(x) is all real numbers except 0 because the function is undefined at x = 0. The range of h(x) is (-∞, 0) U (0, ∞).
The domain and range of the inverse function will be the same as the domain and range of h(x).
d. For the function S(x) = 1/2 (e^x - e^-x), we switch x and y, giving us x = 1/2 (e^y - e^-y). Let's solve for y:
x = 1/2 (e^y - e^-y)
2x = e^y - e^-y
To solve for y, we can add e^-y to both sides of the equation and then take the natural logarithm of both sides:
2x + e^-y = e^y
2x = e^y + e^-y
e^y = 2x + e^-y
e^2y = 2xe^y + 1
e^2y - 2xe^y - 1 = 0
This is a quadratic equation in terms of e^y. We can solve this quadratic equation using the quadratic formula:
e^y = (2x ± sqrt((2x)^2 - 4(1)(-1)))/(2(1))
e^y = (2x ± sqrt(4x^2 + 4))/(2)
e^y = x ± sqrt(x^2 + 1)
To find the inverse, we need to take the natural logarithm of both sides:
y = ln(x ± sqrt(x^2 + 1))
The inverse exists for the entire domain of S(x) because the equation has real solutions for y for any value of x.
The range of S(x) is all real numbers because e^x and e^-x can take all positive and negative real values. The range of S(x) is (-∞, ∞).
The domain and range of the inverse function will be the same as the domain and range of S(x).