Question

Suppose \(f^{\prime}(x)>0\) in \((a, b) .\) Prove that \(f\) is strictly increasing in \((a, b)\), and let \(g\) be its inverse function. Prove that \(g\) is differentiable, and that $$ g^{\prime}(f(x))=\frac{1}{f^{\prime}(x)} \quad(a

Short answer

Question: Prove that if a function \(f\) has a positive derivative (\(f'(x) > 0\)) in an interval \((a,b)\), then the function is strictly increasing in that interval. Furthermore, find the inverse function \(g\) and its derivative, and show that \(g'(f(x)) = \frac{1}{f'(x)}\). Solution: Step 1: Since \(f'(x) > 0\) for all \(x \in (a, b)\), the function \(f\) is strictly increasing in this interval, as the positive slope ensures that the function always increases. Step 2: As \(f\) is strictly increasing in \((a, b)\), it is a bijection and has a unique inverse function \(g\) so that \(g(f(x)) = x\). Step 3: The inverse function theorem states that if \(f\) is differentiable and continuous at a point, and if its derivative is nonzero, then the inverse function is also differentiable and continuous. Since \(f'(x) > 0\) for all \(x \in (a, b)\), the inverse function \(g\) is differentiable in the interval \((f(a), f(b))\). Step 4: By the inverse function theorem, the derivative of the inverse function \(g\) is given by \(g'(f(x)) = \frac{1}{f'(x)}\) for \(a < x < b\).

Step-by-step solution

Prove that f is strictly increasing

Since \(f'(x) > 0\) for all \(x \in (a, b)\), it means that the function \(f\) has a positive slope at every point in the interval \((a, b)\). Thus, if we take any two points \(x_1\) and \(x_2\) such that \(x_1 < x_2\) and both belong to \((a, b)\), \(f(x_1) < f(x_2)\) because the function is always increasing. This proves that \(f\) is strictly increasing in \((a, b)\).

Find the inverse function g

Since \(f\) is strictly increasing in \((a, b)\), it is a bijection, which means it has a unique inverse function \(g\). The inverse function \(g\) maps the values of \(f\) back to their corresponding input values, that is, \(g(f(x)) = x\).

Prove that g is differentiable

As we are given that \(f'(x) > 0\) in \((a, b)\), it is continuous. Therefore, \(f\) is also continuous in its domain. Since \(f\) is a bijection, its inverse function \(g\) exists and is also continuous. Now, we will use the inverse function theorem which states: If \(f\) is differentiable and continuous at a point \(x_0\) in its domain, and if \(f'(x_0) \neq 0\), then its inverse function \(g\) is differentiable and continuous at point \(y_0=f(x_0)\). Since \(f'(x) > 0\) for all \(x \in (a, b)\), the inverse function theorem is applicable, and therefore \(g\) is differentiable in the interval \((f(a), f(b))\).

Find the derivative of g

According to the inverse function theorem, the derivative of \(g\) with respect to its input \(y\) is given by: $$ g'(y) = \frac{1}{f'(g(y))} $$ Now, let's replace \(y\) with \(f(x)\) to get \(g'(f(x))\): $$ g'(f(x)) = \frac{1}{f'(g(f(x)))} $$ Since \(g(f(x)) = x\), the above expression can be simplified to: $$ g'(f(x)) = \frac{1}{f'(x)} \quad(a<x<b) $$ This is the result we were asked to prove.

Key concepts

Strictly Increasing Function

Understanding the nature of a strictly increasing function is crucial when dealing with calculus problems.

A function is described as strictly increasing on an interval if, for any two numbers, let's call them x_1 and x_2, within the interval where x_1 < x_2, the output of the function at x_1 is always less than the output at x_2. In mathematical terms, we write this as: if x_1 < x_2, then f(x_1) < f(x_2).

This property ensures that there are no flat or declining portions within the graph of the function over that interval. It's akin to climbing up a hill that continuously rises; you never encounter a plateau or start descending.

An important consequence of a function being strictly increasing is that it passes the Horizontal Line Test, which implies that a unique inverse function exists. This leads to the question of how the derivative of a function relates to its strictly increasing nature, which can be understood through the fact that if the derivative f'(x) is greater than zero for all points on an interval, the function is strictly increasing on that interval - showing a persistently positive rate of change.

Inverse Function Theorem

The Inverse Function Theorem is a foundational theorem in calculus that addresses the differentiability of inverse functions.

To begin with, what does it mean for a function to have an inverse? An inverse function, symbolized as g when f is the original function, essentially reverses the action of f. It takes the output values of f and converts them back into the original input values, i.e., for any x, g(f(x)) = x. However, not all functions have inverses that are functions themselves; they must be bijective, which means they are both injective (no two inputs have the same output) and surjective (every possible output is produced by some input).

The Inverse Function Theorem comes into play when discussing the differentiability of these inverses. It tells us that if our original function f is both differentiable and its derivative is non-zero at a point x_0, then the inverse g is also differentiable at the point g(x_0). More informally, if you can smoothly trace the graph of f with no jumps, gaps, or vertical tangent lines, and f keeps either increasing or decreasing at the point, then you can assuredly trace the graph of its inverse g smoothly as well.

Differentiability

Differentiability refers to the attribute of a function that allows it to have a derivative at every point within a given interval. For a simplified visual representation, imagine a smooth curve with no sharp corners or interruptions.

When a function is differentiable at a point, it means that the function's graph has a definite tangent line at that point. Subsequently, the function has a certain slope at that point, which is the value of the derivative. If you were to trace your finger along the graph, it would move without jumping or changing direction suddenly.

A critical notion, in line with what was discussed under the Strictly Increasing Function, is that a function with a constant positive derivative will always be increasing, and conversely, a constant negative derivative results in the function always decreasing. Differentiability also has a deep connection with continuity; a function can only be differentiable at points where it is continuous. However, the reverse isn't true - being continuous doesn't guarantee differentiability.

To put it in the context of the original problem, if our initial function f has a positive derivative on the interval (a, b), implying that it's differentiable and hence continues on this interval, it means that not only is f strictly increasing but also that the inverse function g is differentiable. And remarkably, the derivative of the inverse function is the reciprocal of the derivative of the original function evaluated at the inverse function's point - that is, g'(f(x)) = 1 / f'(x).