Question

In Exercises 1-4, find \(\left(f^{-1}\right)^{\prime}(a)\) for the function \(f\) and real number \(a\). $$ \frac{\text { Function }}{f(x)=x^{3}+2 x-1} \quad \frac{\text { Real Number }}{a=2} $$

Short answer

Obtaining the explicit value of \((f^{-1})'(2)\) would depend on the specific numerical methods employed or the software tool used in Step 4. However, the general approach is to find \(f'(x)\) and \(f^{-1}(2)\) and substitute these into the relationship derived in Step 2.

Step-by-step solution

Find inverse function

To find the inverse of the function \(f(x)=x^{3}+2 x-1\), we switch the roles of \(y\) and \(x\). This gives us the equation: \(y = x^{3}+2 x-1\). Solving this equation for \(x\) gives the inverse function which is quite complicated and nontrivial. Note that for the purpose of solving this problem, it is not necessary to find an explicit formula for the inverse function.

Find \((f^{-1})'(a)\)

Although we can't find an explicit formula for the inverse function, we can still find its derivative at a particular point using the formula for the derivative of an inverse function: \((f^{-1})'(a) = 1/(f'(f^{-1}(a)))\).

Calculate \(f'(x)\) and \(f^{-1}(a)\)

We now need to calculate \(f'(x)\) and \(f^{-1}(a)\). The derivative \(f'(x)\) is derived from \(f(x)\) by applying the power rule and gives us: \(f'(x)=3x^{2}+2\).

Solve for \(f^{-1}(2)\)

We know that if \(f(f^{-1}(a)) = a\), then \(f^{-1}(a)\) is a solution of the equation \(f(x)=a\). This means \(f^{-1}(2)\) is a solution of the equation \(f(x)=2\), which simplifies to \(x^{3}+2 x-1=2\). This further simplifies to \(x^{3}+2 x-3=0\), whose solution can be found using a numerical method like Cardano's method or using a software tool.

Calculate \((f^{-1})'(2)\)

Now that we have found \(f'(x)\) and \(f^{-1}(2)\), we can substitute these into the equation from Step 2, yielding \((f^{-1})'(2) = 1/(f'(f^{-1}(2)))\).

Key concepts

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. One of the two major topics in calculus is differential calculus, which concerns the rate at which quantities change, commonly known as the derivative.

When we deal with the derivative of a function, it tells us how a function's output value changes in response to changes in its input value. For example, in physics, the derivative of the position function with respect to time gives us the velocity function, signaling how the position of an object changes over time.

Within calculus, we also encounter specific rules for differentiation, such as the power rule, product rule, and chain rule, which help us to find the derivatives of various types of functions efficiently. Understanding these concepts and how they interconnect is crucial for tackling a wide array of problems in mathematics, physics, engineering, economics, and beyond.

Derivative of Inverse Function

The derivative of an inverse function is an interesting concept that comes up frequently in calculus. If you have a function \(f(x)\) that has an inverse \(f^{-1}(x)\), then the derivative of the inverse function \(\left(f^{-1}\right)^{\prime}(x)\) can be determined using a specific relationship. This relationship is based on the formula \(\left(f^{-1}\right)^{\prime}(a) = 1 / f'\left(f^{-1}(a)\right)\).

To apply this formula correctly, it's crucial to understand each part's roles. Firstly, \(f'\left(f^{-1}(a)\right)\) is the derivative of the original function \(f\) evaluated at the inverse function value that corresponds to \(a\). Then, take the reciprocal of this value to get the derivative of the inverse at \(a\). This step ensures that we consider the interlocked relationship between a function and its inverse.

Importance in Various Fields

Grasping the concept of the derivative of inverse functions becomes particularly important in fields such as mathematics and engineering. For example, if you're looking to find the rate of change of an angle in response to varying arc lengths in trigonometry, you'll likely be working with the derivative of the arc sine function, which is the inverse function of the sine.

Power Rule

The power rule is one of the basic differentiation rules in calculus that allows finding the derivative of a function that can be expressed in the form \(x^n\), where \(n\) is any real number. According to the power rule, \(d/dx(x^n) = nx^{n-1}\).

Applying the power rule makes differentiation quicker and more straightforward -- there is no need for the complex process of limits as one might learn when first introducing derivatives. It is especially useful in polynomials where each term is a power of \(x\), as seen in the original exercise.

For example, in the exercise with the given function \(f(x)=x^{3}+2x-1\), one could quickly find \(f'(x)\) by using the power rule on each term containing \(x\). Adding the derivatives of individual terms, using the power rule for \(x^{3}\) results in \(3x^{2}\), and for \(2x\), the derivative is \(2\). Thus, \(f'(x) = 3x^{2} + 2\), a critically important step in further computations such as finding the derivative of the inverse function at a particular point.