Problem 57 - Calculus Single Variable

Question

Suppose that \(f\) is differentiable at \(c\). Let \(g(x)=f(x+k)\) where \(K\) is a constant. Show that \(g\) is differentiable at \(c-k,\) and evaluate \(g^{\prime}(c-k)\) in terms of \(f^{\prime}(c) .\)

Step-by-Step Solution

Answer

Function \(g(x) = f(x+k)\) is differentiable at \(c-k\), and \(g'(c-k) = f'(c)\).

Step 1: Understanding the Problem

We know that a function is differentiable at a point if its derivative exists at that point. We need to determine whether the function \(g(x) = f(x+k)\) is differentiable at the point \(c-k\), and find \(g'(c-k)\) in terms of \(f'(c)\).

Step 2: Expressing the Derivative of g

First, express \(g'(x)\) using the definition of a derivative: \(g'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h}\). Substitute \(g(x) = f(x+k)\): \[g'(x) = \lim_{h \to 0} \frac{f(x+h+k) - f(x+k)}{h}\]

Step 3: Substituting into the Limit Expression

Now, let's substitute \(x = c-k\) into this limit to find \(g'(c-k)\):\[g'(c-k) = \lim_{h \to 0} \frac{f((c-k)+h+k) - f(c)}{h} = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}\].

Step 4: Relating to the Derivative of f

The expression for \(g'(c-k)\) is exactly the expression for \(f'(c)\), which is the derivative of \(f\) at \(c\): \[g'(c-k) = f'(c)\].

Step 5: Conclusion

Therefore, \(g(x) = f(x+k)\) is differentiable at \(c-k\), and the derivative \(g'(c-k)\) is equal to \(f'(c)\).

Key Concepts

Derivative of a Function

The derivative of a function represents the rate at which the function's value changes as its input changes. For a function to be differentiable at a specific point, the derivative must exist at that point. This means that the slope of the tangent line to the function at that point can be computed.
A derivative is usually denoted as \( f'(x) \), which can be defined using the limit:

• \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \).

This formula gives us a precise understanding of the instantaneous rate of change of the function at any point \( x \).
This concept is crucial in calculus as it allows us to understand how functions behave locally and predict how they change in small time or space intervals.

Chain Rule

The chain rule is a fundamental tool in calculus for finding the derivative of composite functions. It tells us how to differentiate a function that is formed by combining two or more functions.
Suppose you have two functions, \( u(x) \) and \( v(x) \). If a function \( y \) is composed of these two as \( y = u(v(x)) \), then the chain rule helps in determining the derivative of \( y \).

The rule is stated mathematically as:

• \( \frac{dy}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} \).

This formula signifies that the derivative of the outer function \( u \) concerning \( v \) is multiplied by the derivative of \( v \) with respect to \( x \).
The chain rule is widely used in situations where functions are nested within each other and provides a systematic way to compute the derivative of complex expressions.

Function Translation

In mathematics, function translation involves shifting a function horizontally or vertically on a graph. A horizontal translation, specifically, is when every point of the original function moves left or right by some constant value.
A common example of horizontal translation can be described with the formula \( g(x) = f(x+k) \). Here, the graph of \( f(x) \) moves horizontally by \( k \). If \( k \) is positive, the function shifts left; if \( k \) is negative, it shifts right.
When considering differentiability for a translated function like \( g(x) = f(x+k) \), it's essential to account for this shift. The derivative of \( g \) at any point \( x \) considers the same slope as the derivative of \( f \), evaluated at a different point, reflecting the horizontal shift.
Translating functions and understanding their derivatives is crucial for analyzing how operations on functions like shifting affect their behavior, aiding in broader comprehension in calculus.