Question

Use the following tables to determine the indicated derivatives or state that the derivative cannot be determined. $$\begin{array}{lrrrrr}x & -2 & -1 & 0 & 1 & 2 \\\\\hline f(x) & 2 & 3 & 4 & 6 & 7 \\\f^{\prime}(x) & 1 & 1 / 2 &2 & 3 / 2 & 1\end{array}$$ a. \(\left(f^{-1}\right)^{\prime}(4)\) \(\begin{array}{lll}\text { b. }\left(f^{-1}\right)^{\prime}(6) & \text { c. }\left(f^{-1}\right)^{\prime}(1) & \text { d. } f^{\prime}(1)\end{array}\)

Short answer

Question: Determine the values of \((f^{-1})'(4)\), \((f^{-1})'(6)\), \((f^{-1})'(1)\), and \(f'(1)\) using the given data. Answer: a. \((f^{-1})'(4) = \frac{1}{2}\) b. \((f^{-1})'(6) = \frac{2}{3}\) c. Cannot be determined using the given data. d. \(f'(1) = \frac{3}{2}\)

Step-by-step solution

(Step 1: Find \(f^{-1}(4)\))

First, we need to find the value of \(f^{-1}(x)\) for \(x = 4\). From the table, we can see that \(f(0) = 4\). Therefore, \(f^{-1}(4) = 0\).

(Step 2: Find \((f^{-1})'(4)\))

Now, we will use the formula \((f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}\) with \(x = 4\). Since we found that \(f^{-1}(4) = 0\), we can substitute this value into the formula: \((f^{-1})'(4) = \frac{1}{f'(0)}\). From the table, \(f'(0) = 2\). Thus, \((f^{-1})'(4) = \frac{1}{2}\).

(Step 3: Find \(f^{-1}(6)\))

For part b, we need to find \(f^{-1}(x)\) for \(x = 6\). From the table, \(f(1) = 6\). Therefore, \(f^{-1}(6) = 1\).

(Step 4: Find \((f^{-1})'(6)\))

Using the formula \((f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}\) with \(x = 6\), and substituting \(f^{-1}(6) = 1\) into the formula, we get \((f^{-1})'(6) = \frac{1}{f'(1)}\). From the table, \(f'(1) = \frac{3}{2}\). Thus, \((f^{-1})'(6) = \frac{2}{3}\).

(Step 5: Find \(f^{-1}(1)\))

For part c, we need to find \(f^{-1}(x)\) for \(x = 1\). However, looking at the table, we cannot see any corresponding value of \(x\) for which \(f(x) = 1\). Therefore, we cannot determine \((f^{-1})'(1)\) using the given data.

(Step 6: Find \(f'(1)\))

For part d, we are asked to find \(f'(1)\). Looking at the table, we can see that \(f'(1) = \frac{3}{2}\). So, the answers for the given questions are: a. \((f^{-1})'(4) = \frac{1}{2}\). b. \((f^{-1})'(6) = \frac{2}{3}\). c. Cannot be determined using the given data. d. \(f'(1) = \frac{3}{2}\).

Key concepts

Calculus Derivative

Calculus is a branch of mathematics that delves into rates of change. A derivative is a fundamental concept in calculus. It measures how a function’s output changes as its input changes. Think of it as the "slope" of a function at a specific point. Derivatives tell us how a curve changes direction or shape.

When we say we are finding the derivative of a function, we typically use notation like \(f'(x)\) or \(\frac{df}{dx}\). This is crucial for understanding how functions behave, especially when dealing with real-world phenomena like speed or growth.

• Derivatives can be represented graphically as tangent lines on the curve of a function.
• Knowing the derivative lets us predict the behavior of the function in small intervals.
• For complex functions, derivatives can be found using rules like the product rule, quotient rule, and chain rule.

To deepen understanding, you should practice finding derivatives for various functions, which will help make connections between algebraic and graphical interpretations.

Inverse Functions

Inverse functions are like the reverse operations of certain functions. If a function \(f\) takes an input \(x\) and produces an output \(y\), then the inverse function, denoted as \(f^{-1}\), takes \(y\) back to \(x\). Inverse functions "undo" the work of the original function.

To find an inverse function, you switch the roles of the input and output and solve for the new output. Not all functions have inverses, though. A function must be one-to-one (bijective) to have an inverse. This means each output is paired with exactly one input.

• The inverse of a function \(y = f(x)\) is found by solving \(x = f^{-1}(y)\).
• Graphically, an inverse function reflects the original function across the line \(y = x\).

Finding inverses and using them is crucial in calculus when deriving expressions such as inverse derivatives, which are used extensively in optimization problems and differential equations.

Derivative Table Interpretation

Tables are often used to help understand functions and their derivatives without detailed calculations. In these exercises, tables display values of a function and its derivative at specific points. Understanding how to interpret and use these tables helps in solving problems involving inverse derivatives.

Consider a function \(f\) with a given table:

• To find \((f^{-1})'(x)\), use the formula \((f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}\).
• First find \(f^{-1}(x)\) by locating the output in the \(f\) table and noting the corresponding input.
• Then, look up the derivative \(f'\) at the input you found and use it to calculate \((f^{-1})'(x)\).

This approach helps solve problems by quickly referencing values, making it easier to determine derivative values without complex calculations. Practice interpreting these tables to enhance your problem-solving skills in calculus.