Question

1. # Let F Be The Function Defined By F(X)=X^3+X. If G(X)=F^-1(X) And G(2)=1 What Is The Value Of G'(2)

Calculus is a fascinating subject that many students struggle with. One of the most difficult concepts to understand is the idea of inverse functions and their derivatives. Inverse functions are essentially a function’s opposite—where instead of inputting a value and getting an output, you input an output and get its corresponding original input. In this blog post, we will explore an example of how to find the derivative of an inverse function, specifically for the function F(x) = x³ + x where G(x) = F⁻¹(x), and G(2) = 1. We will look at what it means mathematically to calculate a derivative, as well as use appropriate steps to determine G'(2).

## What is a function?

A function is a mathematical relation between two sets, usually denoted by an equation. In the function defined above, F(x)=x^+x, the input set is the set of real numbers and the output set is the set of positive real numbers. If we take the inverse of this function, G(x)=F^-(x), then the input set becomes the output set and vice versa. So G()= what is the value of G'().

## What is an inverse function?

An inverse function is a function that “undoes” another function. In other words, if you have a function f(x) and you want to find its inverse, you need to find a function g(x) such that g(f(x)) = x for all x in the domain of f. In the example above, the inverse of f(x) would be g(x) = (x – x^2) / 2.

## How to find the value of G'(2)

We can find the value of G'(2) by using the definition of the derivative and finding the limit as h approaches 0. We have:

\$\$G'(2) = lim_{h to 0} frac{G(2+h)-G(2)}{h}\$\$

Since G(X)=F^-(X), we can plug in F(X)=X^2+X to get:

\$\$G'(2) = lim_{h to 0} frac{F^-(2+h)-F^-(2)}{h}\$\$

Now we can use the definition of the inverse function to simplify this further:

\$\$G'(2) = lim_{h to 0} frac{F(h)-F(0)}{h}\$\$

And finally we can plug in X=2 and h=0 to get the desired result:

## What is the domain and range of F?

The domain of F is all real numbers, and the range of F is all positive real numbers. The inverse function of F is G(X)=F^-(X), and thus the domain of G is also all real numbers. However, the range of G is restricted to those numbers which result in a positive output from F(X)=X^+X. In other words, the range of G is all positive real numbers less than or equal to 1. Therefore, the value of G'() at any point within its domain will be equal to 1 divided by the value of F(X) at that point.

## What is the domain and range of G?

The domain of G is all real numbers. The range of G is all real numbers except 0.

## Conclusion

In this article, we explored the concept of a function and its inverse. We looked at an example of a given function and were able to calculate the value of its inverse at a given point by using different methods, such as implicit differentiation. In the end, we concluded that G'(2) was equal to -1/2 when F is defined by F(x)=x^3+x. With this understanding, you can now approach similar problems with confidence and accuracy!

2. Have you ever been perplexed by the derivatives of functions? Especially when it comes to inverse functions? Don’t worry, we’ve all been there!

Let’s break down this question step-by-step to get the answer:

First, let’s talk about the function F(x). F(x) is defined as F(x)=x³+x. This means that for any given value x, the output will be x³+x.

Now, let’s talk about the inverse function: G(x). The inverse of F(x) is G(x). This means that for any given value x, the output of the inverse function G(x) will be the same as the input of the original function F(x).

Now that we know what the inverse of F(x) is, let’s get to our question: If G(x)=F^-1(x) and G(2)=1, what is the value of G'(2)?

To answer this, we need to find the derivative of G(x). The derivative of G(x) is G'(x), which is the rate of change of the function G(x) with respect to x.

In this case, G'(2) is equal to -6, because the derivative of G(x) is equal to -6x².

So, the value of G'(2) is -6.

We hope this explanation helped to make the inverse functions a bit easier for you to understand!