f(x) = x^5 + 2x^3 + x - 1
Find f^-1 (3) and (f^-1)' (3)
f^-1 (3) = 1
(f^-1)' (3) = 1/12
Those are the answers, but I have no clue how my teacher got them.
y = x^5 + 2 x^3 + x - 1
exchange x and y
x = y^5 + 2 y^2 + y - 1
now use 3 for x
4 = y^5 + 2 y^2 + y
1 works for y
dy/dx = 5 x^4 + 6 x^2 + 1
but x is 1 here we know already
dy/dx = 5+6 + 1 = 12
interchange
dy/dx = 1/12
Thank You so much!
Actually I do have one question, how do we know that x is 1 already ???
remember 1 worked for y, so not it works for x :)
To find the inverse of a function, denoted by f^⁻1(x), you need to swap the roles of the independent variable (x) and the dependent variable (f(x)). In other words, you need to solve the equation for x instead of f(x). Let's go step by step:
Step 1: Start with the original function f(x) = x^5 + 2x^3 + x - 1.
Step 2: Replace f(x) with "y" to make the equation easier to work with: y = x^5 + 2x^3 + x - 1.
Step 3: Swap x and y to represent the inverse function: x = y^5 + 2y^3 + y - 1.
Step 4: Solve the equation for y. In this case, you need to find the value of y when x = 3.
3 = y^5 + 2y^3 + y - 1.
Step 5: Simplify the equation and solve for y. Since this is a polynomial equation, you may need to use numerical methods or approximate solutions. In this case, the solution is y = 1.
Therefore, f^⁻1(3) = 1.
Now, let's calculate (f^⁻¹)'(3) (the derivative of the inverse function at x = 3). The derivative of the inverse function can be found by using the formula:
(f^⁻¹)'(x) = 1 / f'(f^⁻¹(x)),
where f'(x) is the derivative of the original function f(x).
Step 1: Find the derivative of the original function f(x) = x^5 + 2x^3 + x - 1.
f'(x) = 5x^4 + 6x^2 + 1.
Step 2: Substitute x = f^⁻¹(3) = 1 into the derivative f'(x).
(f^⁻¹)'(3) = 1 / f'(f^⁻¹(3)).
(f^⁻¹)'(3) = 1 / f'(1).
(f^⁻¹)'(3) = 1 / (5(1)^4 + 6(1)^2 + 1).
(f^⁻¹)'(3) = 1 / (5 + 6 + 1).
(f^⁻¹)'(3) = 1 / 12.
Therefore, (f^⁻¹)'(3) = 1/12.
So, the final answers are f^⁻¹(3) = 1, and (f^⁻¹)'(3) = 1/12.