Question

Consider the following functions (on the given internal, if specified). Find the inverse function, express it as a function of \(x,\) and find the derivative of the inverse function. $$f(x)=\sqrt{x+2}, \text { for } x \geq-2$$

Short answer

Question: Find the derivative of the inverse function of \(f(x) = \sqrt{x+2}\), given that the domain of \(x\) is \(x\geq-2\). Answer: \((f^{-1}(x))'=2x\)

Step-by-step solution

Solving for the inverse function

To find the inverse function, we begin by replacing \(f(x)\) with \(y\) for simplification purposes. So, we have: $$y = \sqrt{x+2}$$ Now, switch the roles of \(x\) and \(y\): $$x = \sqrt{y+2}$$ Next, solve for \(y,\) which is our inverse function \(f^{-1}(x).\)

Expressing the inverse function in terms of \(x\)

To solve for \(y,\) we square both sides of the equation to get rid of the square root. $$x^2 = y+2$$ Now, move the constant to the other side of the equation to isolate \(y\): $$y = x^2 -2$$ So, the inverse function is \(f^{-1}(x) = x^2 -2\).

Finding the derivative of the inverse function

Now that we have the inverse function, we need to find its derivative with respect to \(x,\) which we denote as \((f^{-1}(x))'\). To do this, we will apply the power rule and constant rule. $$\frac{d}{dx} (f^{-1}(x)) = \frac{d}{dx} (x^2 - 2)$$ Apply the power rule \(\frac{d}{dx} x^n = nx^{n-1}\) and constant rule \(\frac{d}{dx}k=0\): $$(f^{-1}(x))' = 2x^{2-1} - 0 = 2x^{1}=2x$$ Thus, the derivative of the inverse function is \((f^{-1}(x))'=2x\).

Key concepts

Derivative of Inverse Functions

Understanding the derivative of an inverse function can sometimes be an intimidating task, but it doesn't have to be! The key is to recognize how derivatives of inverse functions are related to the original functions. Given a function and its inverse, the derivative of the inverse function at a particular point is closely tied to the derivative of the original function. This is because the slope of the inverse function is the reciprocal of the slope of the original function at corresponding points. If you know the derivative of the function, say \[ f(x) \], the derivative of its inverse, \[ f^{-1}(y) \], can be calculated using the formula: \[ (f^{-1})'(y) = \frac{1}{f'(x)} \, \text{where} \, x = f^{-1}(y) \]Using this relationship, you can find how quickly values change when moving forward or backward between the functions. Remember not to confuse this with the standard derivative calculation. Instead, this is an exceptional case for inverse functions.

Finding Inverses

Finding the inverse of a function means essentially flipping the roles of the input and output. It's like retracing your steps back to the starting point. To find an inverse:

• First, replace the function notation to make it simpler. Use \(y\) instead of \(f(x)\).
• Next, interchange the variables. So, write \(x = f(y)\), instead of \(y = f(x)\).
• Finally, solve this new equation for \(y\). This isolates \(y\) and gives you the inverse function, \(f^{-1}(x)\).

Remember, not all functions have inverses. Inverses only exist if the function is one-to-one (bijective), meaning each output is paired with exactly one input. A common test for this is the Horizontal Line Test. If a horizontal line crosses your function at more than one point, it doesn't have an inverse. In our original problem, the inverse function of \(f(x) = \sqrt{x+2}\) was found to be \(f^{-1}(x) = x^2 - 2\). This was achieved by following these simple steps.

Power Rule in Calculus

The power rule is one of the most fundamental rules in calculus for finding derivatives. It's especially useful when working with polynomials, which show up frequently in various math problems. The power rule states:For any function of the form \(x^n\), where \(n\) is a constant, the derivative is found as follows:\[ \frac{d}{dx} x^n = n \cdot x^{n-1} \]It's a straightforward operation: you simply bring down the exponent as a multiplier, and then decrease the exponent by one. In our example, the inverse function \(f^{-1}(x) = x^2 - 2\) involves \(x^2\). By applying the power rule, we derive that \((f^{-1}(x))' = 2x\). Note how easily this rule makes differentiating powers of \(x\) manageable. It’s a must-have tool for efficiently navigating more complicated calculus tasks. Add to that, the constant rule applies when differentiating constants (like \(-2\) in our problem), where the derivative is simply zero. The power rule and constant rule together form a perfect toolkit for differentiation.