1. Let f(x)=x^5 + 2x^3 + x - 1
Find f^-1(3) and (f^-1)'(3)?
I have zero idea how to find the inverse of this function at a point 3, and how to take derivative of an inverse.
2.Let f(x)=cosx + 3x
Show that f(x) is a differentiable inverse and find f^-1(1) and (f^-1)'(1).
Inverse again. The answer key says yes, it is differentiable, f^-1(1)=0 and (f^-1)'(1)=1/3.
Thank you so much, I really tried swapping x and y values but I didn't get anywhere with that. Online derivative solver didn't work either. PLease help!
Does f^-1(x) even exist?
f'(x) = 5x^4+6x^2+1
That is always positive, which means that the graph of f(x) does not change direction, so it passes the horizontal-line test.
where is f(x) = 3?
x^5+2x^3+x-1 = 3
x^5+2x^3+x-4 = 0
a little synthetic division shows that x=1 is a solution.
So, f^-1(3) = 1
because f(1) = 3
Google inverse function derivative to find that the slopes of f(x) and f^-1(x) are reciprocals.
f'(3) = 556, so (f^-1)'(3) = 1/556
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f(x) = cosx+3x
f'(x) = 1-sinx
Since f' does not change sign, f has an inverse. Use google for discussions on inverse differentiability.
f(0) = 1, so f^-1(1) = 0.
No problem! I can help you with both of these questions.
1. To find the inverse of a function at a given point, we need to follow these steps:
Step 1: Find the inverse function f^(-1)(x). This involves swapping the x and y variables and solving for y.
To find the inverse of f(x) = x^5 + 2x^3 + x - 1, we can start by replacing f(x) with y to get:
y = x^5 + 2x^3 + x - 1
Next, let's swap the x and y variables:
x = y^5 + 2y^3 + y - 1
Step 2: Solve the equation for y.
This equation may seem difficult to solve for y directly, but we can make use of numerical methods or computer algebra systems to find an approximate solution. One popular numerical method is using the Newton-Raphson method.
However, since you're looking for an exact value for f^(-1)(3), we can use another approach.
First, let's rewrite the equation as a polynomial:
y^5 + 2y^3 + y - (x + 1) = 0
For the given point f^(-1)(3), we need to solve for y when x = 3:
y^5 + 2y^3 + y - 4 = 0
At this point, we can use numerical methods, such as graphing or using a computer algebra system to find approximate solutions. Let's assume that the inverse function is differentiable.
Step 3: Find the derivative of f^(-1)(x).
Once we have the inverse function, f^(-1)(x), we can find its derivative, denoted as (f^(-1))'(x), using the inverse function property:
(f^(-1))'(x) = 1 / f'(f^(-1)(x))
In this case, we want to find (f^(-1))'(3). So, we need to find f^(-1)(3) as our first step, then we can substitute it into the derivative formula to get the answer.
2. To prove that f(x) is a differentiable inverse, we need to show two things:
a) f(x) is a one-to-one function
b) f(x) is continuous and differentiable
For f(x) = cos(x) + 3x, it is indeed a one-to-one function since cosine is periodic and its derivative is always negative or positive, which implies the strict monotonicity except at local extrema.
To find f^(-1)(1), we need to solve the equation f(x) = 1 for x:
1 = cos(x) + 3x
Again, this equation might not have an analytical solution, so we may need to use numerical methods.
Once you find the value of x that satisfies the equation, that will be the value of f^(-1)(1).
To find (f^(-1))'(1), we can use the same derivative formula as in question 1:
(f^(-1))'(x) = 1 / f'(f^(-1)(x))
Then substitute x = 1 into the equation and evaluate it.
Remember, in both cases, finding the inverse function and solving for specific values might involve numerical methods or computer algebra systems to approximate the solutions accurately.