Problem 38 - Precalculus

Question

Determine whether the statement is true or false. Explain your answer. Suppose that \(g(x)=f(x) \sec x,\) where \(f(0)=8\) and \(f^{\prime}(0)=-2 .\) Then $$ \begin{aligned} g^{\prime}(0) &=\lim _{h \rightarrow 0} \frac{f(h) \sec h-f(0)}{h}=\lim _{h \rightarrow 0} \frac{8(\sec h-1)}{h} \\ &=\left.8 \cdot \frac{d}{d x}[\sec x]\right|_{x=0}=8 \sec 0 \tan 0=0 \end{aligned} $$

Step-by-Step Solution

Answer

False, because the calculated derivative is -2, not 0.

Step 1: Interpret the Statement

We need to evaluate whether the given expression for \(g'(0)\) is correct. The function \(g(x)\) is given as \(g(x)=f(x) \sec(x)\), and we have some information about \(f(x)\): \(f(0)=8\) and \(f'(0)=-2\). The statement provides an expression for \(g'(0)\) and claims it is equal to 0.

Step 2: Determine the Derivative of g(x)

The derivative of \(g(x) = f(x) \sec x\) can be found using the product rule: if \(u(x) = f(x)\) and \(v(x) = \sec x\), then \(g'(x) = f'(x) \sec x + f(x) \cdot \frac{d}{dx}(\sec x)\). The derivative of \(\sec x\) is \(\sec x \tan x\). So the expression becomes:\[g'(x) = f'(x) \sec x + f(x) \sec x \tan x\]

Step 3: Evaluate g'(0) Using Given Values

Substitute \(x=0\), \(f(0)=8\), and \(f'(0)=-2\) into the derived expression for \(g'(x)\):\[g'(0) = f'(0) \sec 0 + f(0) \sec 0 \tan 0\]Since \(\sec 0 = 1\) and \(\tan 0 = 0\), we have:\[g'(0) = -2 \cdot 1 + 8 \cdot 1 \cdot 0 = -2\]

Step 4: Compare With the Statement's Expression

The statement gives a different expression for \(g'(0)\), involving limits. It simplifies this expression to imply \(g'(0) = 0\). However, from our calculations, \(g'(0)\) is actually \(-2\). Thus, the statement is inconsistent with our calculation.

Key Concepts

Trigonometric Functions

Trigonometric functions, like sine, cosine, and secant, are foundational in calculus. They often transform and simplify complex expressions. The secant function, denoted as \( \sec x \), is the reciprocal of the cosine function: \( \sec x = \frac{1}{\cos x} \). Understanding secant is crucial when you deal with differentiation involving trigonometric expressions.
The derivative of \( \sec x \) is \( \sec x \tan x \). This formula is essential when you apply the product rule to functions involving secant. In our problem, the derivative of \( \sec x \) plays a vital part in finding the derivative of \( g(x) = f(x) \sec x \).
Recognizing trigonometric derivatives helps differentiate complex functions accurately. With this knowledge, you know how changes in \( x \) affect the behavior of trigonometric functions, such as how they grow or shrink rapidly.

Limits and Continuity

Limits form the backbone of calculus, it is about understanding the behavior of functions as they approach specific points or infinity. In this exercise, the concept of limits is indirectly involved in calculating the derivative of \( g(x) \). The derivative is fundamentally defined as a limit:

• \( g'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} \)

In this scenario, while the statement improperly uses limits to represent \( g'(0) \), understanding limits helps identify and correct this error.
Continuity, closely related to limits, assures us that a function doesn't have any abrupt changes or gaps. For \( g(x) = f(x) \sec x \) to be differentiable, it needs to be continuous at the point \( x = 0 \). Here, the given values for \( f(0) \) and \( f'(0) \) allow us to evaluate the derivative accurately, noting that \( \sec 0 = 1 \), which is continuous.

Calculus Differentiation

Differentiation is about finding how a function changes. The rate of change, or the derivative, provides insights into the slope or the trend of the function here. Product rule, a rule of differentiation, is especially useful when working with products of functions. In our case, the function \( g(x) = f(x) \sec x \) requires the product rule to differentiate.

• The product rule states: \( (uv)' = u'v + uv' \), where \( u = f(x) \) and \( v = \sec x \).

This means we apply the derivative to each function in turn, and then sum the results. Calculating yields \( g'(x) = f'(x) \sec x + f(x) \sec x \tan x \), which highlights the contribution of both functions and their rate of change.
By applying differentiation correctly, we find the error in the original problem, showing that \( g'(0) = -2 \), not 0, which necessitates understanding both trigonometric and differential rules.