Question

In Exercises 5-8, show that the slopes of the graphs of \(f\) and \(f^{-1}\) are reciprocals at the indicated points. $$\begin{array}{ll}\text { Function } & \text { Point } \\\\\hline f(x)=3-4 x & (1,-1)\\\f^{-1}(x)=\frac{3-x}{4} & (-1,1)\end{array}$$

Short answer

The slopes of the graphs of \(f\) and \(f^{-1}\) are reciprocals. The slopes at the given points are -4 and -1/4 respectively.

Step-by-step solution

Compute the derivative of \(f(x)\) and \(f^{-1}(x)\)

We first differentiate \(f(x)=3-4 x\) with respect to \(x\), finding \(f'(x)\). Using the power rule, this gives us \(f'(x)=-4\). \n\nNext, we differentiate the inverse function, \(f^{-1}(x)=\frac{3-x}{4}\) with respect to \(x\), finding \((f^{-1})'(x)\). This gives us \((f^{-1})'(x) = -\frac{1}{4}\).

Evaluate the derivatives at the given points

The slope of the graph of \(f(x)\) at the point \((1,-1)\) is given by \(f'(1)=-4\). Likewise, the slope of the graph of \(f^{-1}(x)\) at the point \((-1,1)\) is given by \((f^{-1})'(-1)=-\frac{1}{4}\).

Show the slopes are reciprocals

We have found that \(f'(1) = -4\) and \((f^{-1})'(-1) = -\frac{1}{4}\). Indeed, these slopes are reciprocals, since \(f'(1) = \frac{1}{(f^{-1})'(-1)}\) and \((f^{-1})'(-1) = \frac{1}{f'(1)}\).

Key concepts

Derivatives of Inverse Functions

Understanding how derivatives operate for inverse functions is vital in calculus. An inverse function essentially 'undoes' the action of the original function. If you have a function denoted as f(x), its inverse, f^-1(x), is the function that delivers an output such that f(f^-1(x)) = x.

When it comes to their derivatives, an interesting relationship emerges. The derivative of an inverse function f^-1(x) at a particular point is the reciprocal of the derivative of the original function f(x) at its corresponding point. In other words, {(f^-1)'} is the reciprocal of f'.

This reciprocal relationship can be expressed mathematically as {(f^-1)'(b) = 1 / f'(a)}, where a = f^-1(b) and b = f(a). The exercise provided perfectly illustrates this concept by showing that the derivative of the function and its inverse at corresponding points are negative reciprocals of each other.

Reciprocal Slopes

The slopes of the graphs of a function and its inverse have a reciprocal relationship at corresponding points. This is deeply connected to the concept of derivatives of inverse functions. Graphically, when a function f is reflected over the line y = x to obtain its inverse f^-1, the slope at any given point on f transforms into a slope on f^-1 such that they are multiplicative inverses.

In our exercise, evaluating the function and its inverse at the given points, we get slopes f'(1) and (f^-1)'(-1), which are -4 and -1/4 respectively. The negative sign indicates direction, while the numerical value shows that the slopes are indeed reciprocals. This reciprocal nature is a fundamental property of the slopes of inverse functions.

Power Rule Differentiation

One of the most fundamental rules in calculus is the power rule for differentiation. It's a quick and efficient way of finding the derivative of a function where the variable x is raised to a power. The power rule states that if f(x) = xⁿ, then the derivative f'(x) = nx^n-1.

This rule is elegantly simple and can be applied to any real number power n. It was utilized in the exercise when differentiating f(x) = 3 - 4x, treating -4x as -4x¹, leading to a derivative of f'(x) = -4. By applying this rule, we were able to quickly find the slope of both the function and its inverse, enabling us to verify their reciprocal nature.

• For f(x) = xⁿ, the derivative is f'(x) = nx^n-1.
• The power rule is applicable to both positive and negative powers of x, except when x = 0.

Having a firm grasp of the power rule simplifies many calculus problems and is a key component of understanding derivatives, including those of inverse functions.