Problem 122
Question
Given that \(f^{\prime}(x)>g^{\prime}(x)\) for all real \(x\) and \(f(0)=g(0)\),
then
(A) \(f(x)>g(x) \forall x \in(0, \infty)\)
(B) \(f(x)
Step-by-Step Solution
Verified Answer
(A) and (D) are correct; \(f(x) > g(x)\) for all real \(x\).
1Step 1: Understand the Problem
We are given two derivative functions, \(f'(x)\) and \(g'(x)\), with \(f'(x) > g'(x)\) for all real \(x\). This implies that \(f(x)\) is increasing at a greater rate than \(g(x)\) everywhere on the real line. Both functions have the same starting point at \(x = 0\), i.e., \(f(0) = g(0)\).
2Step 2: Compare Function Values for Positive x
Since \(f'(x) > g'(x)\), it follows that \(f(x)\) is increasing faster than \(g(x)\), resulting in \(f(x) > g(x)\) for \(x > 0\). Therefore, \(f(x) > g(x)\) in the interval \((0, \infty)\).
3Step 3: Analyze Effect on Negative x
For \(x < 0\), \(f'(x) > g'(x)\) still holds, so the rate at which \(f(x)\) increases (or decreases less) is greater than that of \(g(x)\). Below zero, while both might be decreasing, \(f(x)\) decreases at a slower rate than \(g(x)\), leading to \(f(x) > g(x)\). Thus, \(f(x) > g(x)\) for \(x < 0\).
4Step 4: Validate the Full Domain Comparisons
Given that \(f(x) > g(x)\) in both specified intervals \((0, \infty)\) and \((-\infty, 0)\), options (A) and (D) are both valid. However, the problem asks for a consistent result given all and not distinguishing by sign of \(x\), indicating (A) is the most aligned.
Key Concepts
Derivative ComparisonFunction Behavior AnalysisInterval Analysis
Derivative Comparison
When dealing with functions characterized by their derivatives, a significant focus is on understanding how the derivatives of these functions compare to one another. Given the exercise problem, where we have two derivatives, \(f'(x)\) and \(g'(x)\), and it's stated that \(f'(x) > g'(x)\) for all real numbers \(x\), this implies that \(f(x)\) is increasing at a greater rate than \(g(x)\).
The derivative of a function tells us the rate of change or the slope of the function at any given point along the curve. To say that \(f'(x) > g'(x)\) is to observe that at every point \(x\), the slope of \(f(x)\) is steeper in the increasing direction than that of \(g(x)\).
The derivative of a function tells us the rate of change or the slope of the function at any given point along the curve. To say that \(f'(x) > g'(x)\) is to observe that at every point \(x\), the slope of \(f(x)\) is steeper in the increasing direction than that of \(g(x)\).
- If \(f'(x) > g'(x)\), then \(f(x)\) rises faster than \(g(x)\).
- The consequence of this is important when evaluating how the function values of \(f\) and \(g\) compare as \(x\) changes.
- Such a comparison gives us insights into their behavior over the entire real axis.
Function Behavior Analysis
Function behavior involves understanding how functions themselves evolve over the domain in response to their derivatives. In this context, where \(f'(x) > g'(x)\) consistently, we are led to conclude certain things about their behaviors over specific intervals.
When analyzing the interval \((0, \infty)\), since \(f(0) = g(0)\) and \(f'(x) > g'(x)\) for \(x > 0\), it follows that \(f(x) > g(x)\) as \(f(x)\) continues to grow faster. Essentially, both functions start from the same point, but \(f(x)\)'s steeper ascent ensures it outpaces \(g(x)\).
As for the interval \((-\infty, 0)\), although potentially both functions may be decreasing, the slower decrease in \(f(x)\) compared to \(g(x)\) (since \(f'(x) > g'(x)\)) leads again to \(f(x) > g(x)\) throughout this interval. This implies that even when shrinking, \(f(x)\) maintains an advantage over \(g(x)\).
When analyzing the interval \((0, \infty)\), since \(f(0) = g(0)\) and \(f'(x) > g'(x)\) for \(x > 0\), it follows that \(f(x) > g(x)\) as \(f(x)\) continues to grow faster. Essentially, both functions start from the same point, but \(f(x)\)'s steeper ascent ensures it outpaces \(g(x)\).
As for the interval \((-\infty, 0)\), although potentially both functions may be decreasing, the slower decrease in \(f(x)\) compared to \(g(x)\) (since \(f'(x) > g'(x)\)) leads again to \(f(x) > g(x)\) throughout this interval. This implies that even when shrinking, \(f(x)\) maintains an advantage over \(g(x)\).
- Analysis focuses on how comparative changes in rate affect the overall behavior of function values.
- Starting conditions such as \(f(0) = g(0)\) establish the baseline for further analysis.
- The real influence lies in the persistent greater derivative of \(f(x)\).
Interval Analysis
In calculus problems like this one, interval analysis helps determine where certain inequalities hold true for functions based on their derivatives. Here, we consider different segments of the real line where \(f'(x) > g'(x)\).
For the positive interval \((0, \infty)\), since \(f(x)\) and \(g(x)\) start equal at the origin and \(f(x)\) increases more quickly, \(f(x)\) will always be greater than \(g(x)\) as \(x\) increases. This concludes that \(f(x) > g(x)\) for positive \(x\).
In the negative interval \((-\infty, 0)\), the same derivative relationship signifies \(f(x)\) reduces less dramatically compared to \(g(x)\), maintaining \(f(x)\)'s superiority over \(g(x)\) even as they both might decrease.
For the positive interval \((0, \infty)\), since \(f(x)\) and \(g(x)\) start equal at the origin and \(f(x)\) increases more quickly, \(f(x)\) will always be greater than \(g(x)\) as \(x\) increases. This concludes that \(f(x) > g(x)\) for positive \(x\).
In the negative interval \((-\infty, 0)\), the same derivative relationship signifies \(f(x)\) reduces less dramatically compared to \(g(x)\), maintaining \(f(x)\)'s superiority over \(g(x)\) even as they both might decrease.
- By understanding which interval each statement applies to, one can more easily conclude the overall function relationship.
- Intervals dictate the manner in which derivative information translates to function value comparisons.
- The consistent derivative inequality is the foundation for these interval-based conclusions.
Other exercises in this chapter
Problem 120
The function \(f(x)=\frac{|x-1|}{x^{2}}\) (A) increases in \((-\infty, 0) \cup(1,2)\) (B) increases in \((0,1) \cup(2, \infty)\) (C) decreases in \((0,1) \cup(2
View solution Problem 121
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View solution Problem 123
For the function \(f(x)=\int_{2}^{x} e^{-t^{4} / 4}\left(4-t^{2}\right) d t\), (A) maximum occurs at \(x=2\) (B) minimum occurs at \(x=-2\) (C) maximum occurs a
View solution Problem 124
Let \(f(x)=\sin x+\frac{1}{2} \cos 2 x\). Then (A) \(\min\) \(x \in\left[0, \frac{\pi}{2}\right] f(x)\frac{3}{4}}\) (C) \(\min _{x \in\left[0, \frac{\pi}{2}\rig
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