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Chapter 3: Problem 48

Consider the following functions. In each case, without finding the inverse, evaluate the derivative of the inverse at the given point. $$f(x)=\frac{1}{2} x+8 ;(10,4)$$

Short Answer

Expert verified
Solution: The derivative of the inverse function at the point (10, 4) is \(2\).

Step by step solution

01

Find the derivative of f(x)

To find the derivative of \(f(x) = \frac{1}{2}x + 8\), we simply differentiate it with respect to x: $$f'(x) = \frac{1}{2}$$
02

Find the inverse function value at the given point

We know that the given point \((10, 4)\) lies on the inverse function. This means that: $$f(4) = 10$$ Now, we need to determine the value of the inverse function \(g(10)\), which is the x-coordinate of the given point. Since we know that \(f(g(x)) = x\), we substitute the values we have: $$f(4) = \frac{1}{2}(g(10)) + 8$$ Solving for \(g(10)\): $$10 = \frac{1}{2}(g(10)) + 8$$ $$g(10) = 4$$
03

Apply the derivative of the inverse function formula

Now that we have the values for \(f'(x)\) and \(g(x)\), we can substitute them into the formula for the derivative of the inverse function: $$g'(10) = \frac{1}{f'(g(10))} = \frac{1}{f'(4)}$$ Now, replace \(f'(4)\) with the constant value we found in Step 1: $$g'(10) = \frac{1}{\frac{1}{2}}$$
04

Evaluate the derivative of the inverse function

Finally, we can evaluate the derivative of the inverse function at the given point: $$g'(10) = 2$$ So, the derivative of the inverse function \(g'(x)\) at the point \((10, 4)\) is equal to \(2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative
The derivative of a function is a fundamental concept in calculus. It represents the rate at which a function changes at any given point. This is commonly referred to as the slope of the function's graph at that point. In simple terms, it's how steep the curve is. The derivative is denoted by \( f'(x) \) for a function \( f(x) \). In the provided exercise, the derivative of the function \( f(x) = \frac{1}{2}x + 8 \) is constant because the function is linear. Hence, \( f'(x) = \frac{1}{2} \). This tells us the function's graph has a constant slope of \( \frac{1}{2} \), meaning it's a straight line inclined gently upwards.
Function Evaluation
Function evaluation involves finding the output of a function for a specific input. In other words, you're determining \( f(x) \) for a particular \( x \). This step confirms whether the point is part of the function. For instance, in the exercise, the pair \((10, 4)\) indicates a relationship between the function and its inverse. If \( (10, 4) \) is a point on the inverse, \( f(4) \) must equal 10, aligning the function and its inverse correctly. Function evaluation helps to solidify the connection between \( f(x) \) and its inverse, making it essential for verifying such pairs. It simplifies understanding whether a point truly belongs to a function or its inverse.
Inverse Function Derivative
The derivative of an inverse function provides insight into how an inverse function changes. To find this derivative, we use a special formula: \( g'(x) = \frac{1}{f'(g(x))} \). This formula offers a useful way to evaluate the derivative without directly finding the inverse function. In our example, since \( f'(x) = \frac{1}{2} \), and we've determined \( g(10) = 4 \), we calculated the derivative \( g'(10) \) as \( g'(10) = \frac{1}{f'(4)} = 2 \). This calculation highlights that the change in the inverse function at this point is twice as fast as its original function's slope, illustrating the reflective nature of inverse functions in calculus.
Calculus
Calculus is the branch of mathematics focused on change and motion. It has two foundational areas: differentiation, which deals with rates of change, and integration, which handles accumulation of quantities. In this exercise, differentiation is key, especially in understanding how derivatives relate to inverse functions. Calculus allows you to explore the behavior of functions and their inverses through derivatives and rates of change. This understanding is crucial when considering slopes, curves, and the nature of continuous change. By studying calculus, students gain tools to analyze complex dynamics in both mathematics and real-world applications.

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Most popular questions from this chapter

The lapse rate is the rate at which the temperature in Earth's atmosphere decreases with altitude. For example, a lapse rate of \(6.5^{\circ}\) Celsius / km means the temperature decreases at a rate of \(6.5^{\circ} \mathrm{C}\) per kilometer of altitude. The lapse rate varies with location and with other variables such as humidity. However, at a given time and location, the lapse rate is often nearly constant in the first 10 kilometers of the atmosphere. A radiosonde (weather balloon) is released from Earth's surface, and its altitude (measured in kilometers above sea level) at various times (measured in hours) is given in the table below. $$\begin{array}{lllllll} \hline \text { Time (hr) } & 0 & 0.5 & 1 & 1.5 & 2 & 2.5 \\ \text { Altitude (km) } & 0.5 & 1.2 & 1.7 & 2.1 & 2.5 & 2.9 \\ \hline \end{array}$$ a. Assuming a lapse rate of \(6.5^{\circ} \mathrm{C} / \mathrm{km},\) what is the approximate rate of change of the temperature with respect to time as the balloon rises 1.5 hours into the flight? Specify the units of your result and use a forward difference quotient when estimating the required derivative. b. How does an increase in lapse rate change your answer in part (a)? c. Is it necessary to know the actual temperature to carry out the calculation in part (a)? Explain.

Use implicit differentiation to find\(\frac{d y}{d x}.\) $$x+2 y=\sqrt{y}$$

a. Differentiate both sides of the identity \(\cos 2 t=\cos ^{2} t-\sin ^{2} t\) to prove that \(\sin 2 t=2 \sin t \cos t\) b. Verify that you obtain the same identity for \(\sin 2 t\) as in part (a) if you differentiate the identity \(\cos 2 t=2 \cos ^{2} t-1\) c. Differentiate both sides of the identity \(\sin 2 t=2 \sin t \cos t\) to prove that \(\cos 2 t=\cos ^{2} t-\sin ^{2} t\)

Work carefully Proceed with caution when using implicit differentiation to find points at which a curve has a specified slope. For the following curves, find the points on the curve (if they exist) at which the tangent line is horizontal or vertical. Once you have found possible points, make sure that they actually lie on the curve. Confirm your results with a graph. $$y^{2}-3 x y=2$$

The following equations implicitly define one or more functions. a. Find \(\frac{d y}{d x}\) using implicit differentiation. b. Solve the given equation for \(y\) to identify the implicitly defined functions \(y=f_{1}(x), y=f_{2}(x), \ldots.\) c. Use the functions found in part (b) to graph the given equation. \(x+y^{3}-x y=1\) (Hint: Rewrite as \(y^{3}-1=x y-x\) and then factor both sides.)
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