Problem 2 - Calculus - AP Edition

Question

Let \(f(x)=x^{3}+2 e^{x}\) (a) Show that \(f\) is one-to-one and confirm that \(f(0)=2\) (b) Find \(\left(f^{-1}\right)^{\prime}(2)\)

Step-by-Step Solution

Answer

(a) The function is one-to-one and \(f(0)=2\). (b) \((f^{-1})'(2) = \frac{1}{2}\).

Step 1: Prove that function is one-to-one

To show that the function \( f(x) = x^3 + 2 e^x \) is one-to-one, we need to show that its derivative is always positive or always negative. Calculate the derivative: \[ f'(x) = 3x^2 + 2e^x \]Both \(3x^2\) and \(2e^x\) are non-negative for all \(x\), and \(2e^x\) is always positive. Hence, \(f'(x) > 0\) for all \(x\), so \(f(x)\) is strictly increasing and thus one-to-one.

Step 2: Confirm value at given point

Next, we need to confirm that \( f(0) = 2 \). Substitute \(x = 0\) into the function:\[ f(0) = 0^3 + 2e^0 = 0 + 2 \cdot 1 = 2 \]Therefore, \( f(0) = 2 \) is confirmed.

Step 3: Understanding the inverse derivative

The formula to find the derivative of the inverse function at a point \(b\) is:\[(f^{-1})'(b) = \frac{1}{f'(a)}\] where \(f(a) = b\). We need to find \((f^{-1})'(2)\) implying we need \(a\) such that \(f(a) = 2\). Since we have already shown \(f(0) = 2\), our \(a = 0\).

Step 4: Calculate the inverse derivative at the point

Using the formula:\[(f^{-1})'(2) = \frac{1}{f'(0)}\]Calculate \(f'(0)\):\[f'(0) = 3 \cdot 0^2 + 2e^0 = 2 \cdot 1 = 2\]Thus:\[(f^{-1})'(2) = \frac{1}{2}\]

Key Concepts

One-to-One Functions

In mathematics, a function is described as one-to-one (or injective) if each output value in the function's range is linked to by exactly one input value from the domain. This simply means that no two different inputs will ever produce the same output.Identifying these functions is crucial because they guarantee that the function has an inverse that is also a function. To demonstrate if a function is one-to-one, you can utilize one of the following methods:

• Algebraically: Show that if \(f(a) = f(b)\), then it must be that \(a = b\). This approach can sometimes be laborious depending on the complexity of the function.
• Using Derivatives: For functions that are differentiable, you can find the derivative \(f'(x)\). If \(f'(x)\) is always positive or always negative across its domain, then the function is strictly monotonic and therefore, one-to-one.

For the function \(f(x) = x^3 + 2e^x\), the derivative \(f'(x) = 3x^2 + 2e^x\) is always positive since each term \(3x^2\) and \(2e^x\) is non-negative and \(2e^x\) is strictly positive. This ensures that \(f(x)\) is always increasing, confirming that it is one-to-one.

Derivatives

Derivatives represent the rate at which a function is changing at any given point and are a fundamental tool in calculus. By differentiating a function, you can find information about its behavior, such as where it increases or decreases, and identify any local extrema (maximum or minimum points).For instance, if a derivative is always positive over an interval, the function is rising over that interval. Thus, for the function \(f(x) = x^3 + 2e^x\), the derivative \(f'(x) = 3x^2 + 2e^x\) tells us that the function is strictly increasing.Here’s why the derivative \(f'(x)\) works this way:

• Term "\(3x^2\)": This is always non-negative, as \(x^2\) squares the input value, making it zero or positive regardless of whether \(x\) is positive or negative.
• Term "\(2e^x\)": This is always positive because the exponential function \(e^x\) never equals zero and grows rapidly as \(x\) increases.

So, the derivative \(f'(x) = 3x^2 + 2e^x\) is positive for all \(x\). This directly indicates that the original function \(f(x)\) is one-to-one, thanks to its consistently increasing nature.

Inverse Functions

An inverse function basically "reverses" the output of the original function to retrieve the input. If a function \(f\) is one-to-one, it will have an inverse, denoted as \(f^{-1}\).Calculating the derivative of the inverse function is a handy technique in calculus, particularly when you need to compute slopes of lines tangent to graphs of inverse functions. The relationship between the derivative of a function and its inverse is given by:\[(f^{-1})'(b) = \frac{1}{f'(a)}\]Here, \(b\) is the value for which you're calculating the inverse derivative, and \(a\) is the function's input that provides \(b\) when plugged into \(f\). In simpler terms, this means to find \((f^{-1})'(b)\), you need the point \(a\) where \(f(a) = b\).As an illustration: when finding \((f^{-1})'(2)\) for \(f(x) = x^3 + 2e^x\), since \(f(0) = 2\), we use:\[(f^{-1})'(2) = \frac{1}{f'(0)}\]With \(f'(0) = 2\) (calculated through \(3 \cdot 0^2 + 2e^0\)), we conclude:\[(f^{-1})'(2) = \frac{1}{2}\]This result provides the derivative of the inverse function at the point 2, offering insight into its behavior around this specific value.