/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 Consider the following functions... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the following functions (on the given interval, if specified). Find the derivative of the inverse function. $$f(x)=x^{3}+3$$

Short Answer

Expert verified
Answer: The derivative of the inverse function is $$(f^{-1}(x))' = \frac{1}{3}(x - 3)^{-\frac{2}{3}}$$.

Step by step solution

01

Find the inverse function

To find the inverse function, we will first replace $$f(x)$$ with $$y$$, and then swap $$x$$ and $$y$$ in the equation. Finally, we will solve for the new $$y$$ to obtain the inverse function. Let's do that: 1. Replace $$f(x)$$ with $$y$$: $$y = x^3 + 3$$ 2. Swap $$x$$ and $$y$$: $$x = y^3 + 3$$ 3. Solve for the new $$y$$: $$y^3 = x - 3$$ $$y = \sqrt[3]{x - 3}$$ The inverse function is $$f^{-1}(x) = \sqrt[3]{x - 3}$$.
02

Differentiate the inverse function

To differentiate the inverse function $$f^{-1}(x) = \sqrt[3]{x - 3}$$, we will use the chain rule for differentiation. The process is as follows: 1. Write the inverse function as a power function: $$f^{-1}(x) = (x - 3)^{\frac{1}{3}}$$ 2. Apply the chain rule: $$(f^{-1}(x))' = \frac{1}{3} (x - 3)^{\frac{1}{3} - 1} \cdot (\frac{d}{dx}(x - 3))$$ 3. Differentiate the inside function \((x - 3)\): $$\frac{d}{dx}(x - 3) = 1$$ 4. Substitute this back in and simplify: $$(f^{-1}(x))' = \frac{1}{3} (x - 3)^{\frac{1}{3} - 1} \cdot 1$$ $$(f^{-1}(x))' = \frac{1}{3}(x - 3)^{-\frac{2}{3}}$$ The derivative of the inverse function is $$(f^{-1}(x))' = \frac{1}{3}(x - 3)^{-\frac{2}{3}}$$.

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Key Concepts

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

Inverse Functions
When we talk about inverse functions, we are referring to a pair of functions that effectively 'undo' each other. To visualize this, think of a simple operation like addition and its inverse, subtraction. For a function to have an inverse, it must be a bijective function, meaning it is both injective (one-to-one) and surjective (onto).

In mathematical terms, given a function f, if applying f to an input x gives us a result y (written as f(x) = y), then applying the inverse function, denoted as f-1, to y should give us back x (f-1(y) = x). The process of finding the inverse involves swapping the roles of the dependent and independent variables, then solving for the new dependent variable.

To guarantee that every input has a unique output (which is necessary for a function to have an inverse), we sometimes restrict the domain of the original function. For example, the square function, x2, does not have an inverse on all real numbers since it is not one-to-one; however, if we restrict its domain to non-negative numbers, the inverse function becomes the square root function.
Differentiation
Differentiation is a cornerstone concept in calculus that deals with finding the rate at which a quantity changes. In simpler terms, when we differentiate a function, we are finding the derivative, which tells us the slope of the function at any point.

The derivative is a powerful tool because it provides instantaneous rates of change and can be understood as the slope of the tangent line to the curve of a function at a given point. This concept applies to a wide array of physical phenomena, such as velocity, which is the rate of change of distance with respect to time.

Mathematically, the derivative of a function at a point is defined as the limit of the average rate of change of the function over an interval as the interval becomes infinitesimally small. This limit, if it exists, either gives us the exact rate of change at a specific point (if we are dealing with a smooth curve) or it can be undefined (if the function has a sharp corner or discontinuity at the point).
Chain Rule
The chain rule is a fundamental theorem in calculus used for differentiating composite functions. In other words, when one function is nested within another, the chain rule enables us to find the derivative of this compound function.

To apply the chain rule, we need to identify the 'outer' function and the 'inner' function. If we have a function h(x) that can be written as g(f(x)), where f is the inner and g is the outer function, then the chain rule tells us that the derivative of h with respect to x is the derivative of g with respect to f(x), multiplied by the derivative of f with respect to x. Mathematically, it's expressed as h'(x) = g'(f(x)) · f'(x).

The chain rule is especially useful when dealing with functions raised to powers or functions involving roots, as we saw in the textbook solution example. By recognizing the composition of functions, the chain rule allows us to systematically break down the differentiation process into manageable steps, making it easier to find the derivative of complex expressions.

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Most popular questions from this chapter

The total energy in megawatt-hr (MWh) used by a town is given by $$E(t)=400 t+\frac{2400}{\pi} \sin \frac{\pi t}{12}$$ where \(t \geq 0\) is measured in hours, with \(t=0\) corresponding to noon. a. Find the power, or rate of energy consumption, \(P(t)=E^{\prime}(t)\) in units of megawatts (MW). b. At what time of day is the rate of energy consumption a maximum? What is the power at that time of day? c. At what time of day is the rate of energy consumption a minimum? What is the power at that time of day? d. Sketch a graph of the power function reflecting the times when energy use is a minimum or a maximum.

Use implicit differentiation to find\(\frac{d y}{d x}.\) $$6 x^{3}+7 y^{3}=13 x y$$

a. Determine an equation of the tangent line and the normal line at the given point \(\left(x_{0}, y_{0}\right)\) on the following curves. (See instructions for Exercises 73-78. b. Graph the tangent and normal lines on the given graph. \(\left(x^{2}+y^{2}\right)^{2}=\frac{25}{3}\left(x^{2}-y^{2}\right); \left(x_{0}, y_{0}\right)=(2,-1)\) (lemniscate of Bernoulli) (Graph cant copy)

Vibrations of a spring Suppose an object of mass \(m\) is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position \(y=0\) when the mass hangs at rest. Suppose you push the mass to a position \(y_{0}\) units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system , the position y of the mass after t seconds is $$y=y_{0} \cos (t \sqrt{\frac{k}{m}})(4)$$ where \(k>0\) is a constant measuring the stiffness of the spring (the larger the value of \(k\), the stiffer the spring) and \(y\) is positive in the upward direction. A mechanical oscillator (such as a mass on a spring or a pendulum) subject to frictional forces satisfies the equation (called a differential equation) $$y^{\prime \prime}(t)+2 y^{\prime}(t)+5 y(t)=0$$ where \(y\) is the displacement of the oscillator from its equilibrium position. Verify by substitution that the function \(y(t)=e^{-t}(\sin 2 t-2 \cos 2 t)\) satisfies this equation.

Vibrations of a spring Suppose an object of mass \(m\) is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position \(y=0\) when the mass hangs at rest. Suppose you push the mass to a position \(y_{0}\) units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system , the position y of the mass after t seconds is $$y=y_{0} \cos (t \sqrt{\frac{k}{m}})(4)$$ where \(k>0\) is a constant measuring the stiffness of the spring (the larger the value of \(k\), the stiffer the spring) and \(y\) is positive in the upward direction. Use equation (4) to answer the following questions. a. The period \(T\) is the time required by the mass to complete one oscillation. Show that \(T=2 \pi \sqrt{\frac{m}{k}}\) b. Assume \(k\) is constant and calculate \(\frac{d T}{d m}\) c. Give a physical explanation of why \(\frac{d T}{d m}\) is positive.

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