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Chapter 2: Problem 14

If \(f\) is odd, which of the following is right hand derivative of \(f\) at \(x=-a ?\) (a) \(\lim _{h \rightarrow 0^{-}} \frac{f(a-h)-f(a)}{-h}\) (b) \(\lim _{h \rightarrow 0^{-}} \frac{f(h-a)-f(a)}{h}\) (c) \(\lim _{h \rightarrow 0^{+}} \frac{f(a)+f(a-h)}{-h}\) (d) \(\lim _{k \rightarrow 0^{-}} \frac{f(-a)-f(-a-h)}{-h}\)

Short Answer

Expert verified
The short answer is: (d) \(\lim _{h \rightarrow 0^{-}} \frac{f(-a)-f(-a-h)}{-h}\).

Step by step solution

01

Definition of an Odd Function

An odd function is a function that satisfies the condition f(-x) = -f(x) for all values of x in its domain. When we know that the function f is odd, we can use this property to find the value of f at -x when we know the value at x.
02

Definition of the Right Hand Derivative

The right hand derivative of a function f at point x=a is the limit of the expression \(\frac{f(a+h)-f(a)}{h}\), as h approaches 0 from the positive side. In mathematical notation: \[ \text{Right hand derivative of } f \text{ at } x=a \: = \: \lim_{h \rightarrow 0+} \frac{f(a+h)-f(a)}{h} \]
03

Find the Expression of the Right Hand Derivative at x=-a

We need to find the right hand derivative of f at x = -a. Using the definition from Step 2, we can substitute -a for a in the expression: \[ \lim_{h \rightarrow 0^{+}} \frac{f(-a+h)-f(-a)}{h} \]
04

Apply the Odd Function Property

Recall from Step 1 that the odd function property states that f(-x) = -f(x). We'll apply this property to the previous expression: \[ \lim_{h \rightarrow 0^{+}} \frac{-f(a-h)+f(a)}{h} \]
05

Rewrite Expression with Negative Sign

We can rewrite the expression to incorporate the negative sign in the denominator and simplify the result: \[ \lim_{h \rightarrow 0^{+}} \frac{f(a)-f(a-h)}{-h} \]
06

Compare the Result with Given Options

Comparing our final expression with the given options, we can see that the correct answer is: (d) \(\lim _{h \rightarrow 0^{-}} \frac{f(-a)-f(-a-h)}{-h}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Odd Function
An odd function is a fascinating concept in mathematics. By definition, a function is considered odd if it satisfies the condition that, for every point in its domain, the function value at the negative of that point is equal to the negative of the function value at the point itself:
  • Mathematically, this is written as: \( f(-x) = -f(x) \).
  • Odd functions have a peculiar symmetry. They are symmetric about the origin on a graph. This means that if you rotate the graph 180 degrees around the origin, it would look the same.
  • Examples of odd functions include \( f(x) = x^3 \) and \( f(x) = \sin(x) \).
Understanding odd functions is useful when solving calculus problems, as their symmetrical nature can simplify the differentiation process.
Limit Definition of Derivative
The limit definition of a derivative is foundational in calculus. It allows us to determine the rate at which a function changes at a given point. This change is captured by the concept of a derivative, which is fundamentally a limit:
  • The derivative of a function \( f \) at a point \( x = a \) is given by: \[\lim_{h \rightarrow 0} \frac{f(a+h) - f(a)}{h}.\]
  • The expression uses the increment \( h \), which approaches zero, to calculate the instantaneous rate of change at the point \( a \).
  • This foundational concept opens doors to understanding more complex calculus problems, such as finding tangents, velocities, and other real-world applications.
Grasping this definition is key to analyzing how different functions behave at specific points.
Derivative at a Point
Finding the derivative of a function at a specific point reveals how the function behaves exactly at that location. The rate of change can be thought of as the slope of the tangent line to the graph at that point:
  • When dealing with the right-hand derivative, the approach is slightly different. We observe the function's behavior as we approach from the right side of the point.
  • The right-hand derivative at \( x = a \) is calculated as \[\lim_{h \rightarrow 0^+} \frac{f(a+h) - f(a)}{h}. \]
  • This gives insights into immediate changes as we move just to the right of the point. It's especially useful in situations involving right-hand continuity or one-sided limits.
Mastering how to find the derivative at a point is a crucial skill for tackling diverse calculus problems.
Calculus Problem Solving
Solving calculus problems often involves understanding the function, its properties, and how it behaves under different conditions. Here are some strategies to approach calculus problems:
  • Begin by identifying what type of function you're working with (e.g., odd, even) and the specific property you need to apply.
  • Use the limit definition of the derivative as a central tool for finding how the function changes at any given point.
  • Break down the problem into smaller steps. Often, you'll solve part of the problem then apply that knowledge in subsequent steps.
  • Compare your findings to given options or expected outcomes to see if your results align with them.
Consistency and practice in solving various types of calculus problems will build a strong foundation for advanced mathematical challenges.

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Most popular questions from this chapter

Let \([x]\) denotes the integral part of \(x, g(x)=x-[x] .\) Let \(f(x)\) be any continuous function with \(f(0)=f(1)\), then the function \(h(x)=f(g(x))\) : (a) has finitely many discontinuities (b) is discontinuous at some \(x=c\) (c) is continuous on \(\mathbb{R}\) (d) is a constant function

Discuss the continuity on \(0 \leq x \leq 1\) and differentiability at \(x=0\) for the function. \(f(x)=x \cdot \sin \frac{1}{x} \cdot \sin \frac{1}{x \cdot \sin \frac{1}{x}}\) where \(x \neq 0, x \neq 1 / r \pi\) and \(f(0)=f(1 / r \pi)=0, r=1,2,3, \ldots \ldots\)

If \(f(x)=\cos x+|\cos x|\), then evaluate \(\left|f^{\prime}\left(\frac{\pi^{-}}{2}\right)+f^{\prime}\left(\frac{\pi^{+}}{2}\right)\right| .\)

For \(x>0\), let \(h(x)=\left\\{\begin{array}{ll}\frac{1}{q} & \text { if } x=\frac{p}{q} \\ 0 & \text { if } x \text { is irrational }\end{array} ;\right.\) where \(p\) and \(q>0\) are relatively prime integers then which one does not hold good? (a) \(h(x)\) is discontinuous for all \(x\) in \((0, \infty)\) (b) \(h(x)\) is continuous for each irrational in \((0, \infty)\) (c) \(h(x)\) is discontinuous for each rational in \((0, \infty)\) (d) \(h(x)\) is not derivable for all \(x\) in \((0, \infty)\).

Indicate all correct alternatives if, \(f(x)=\frac{x}{2}-1\), then on the interval \([0, \pi]\) (a) \(\tan (f(x))\) and \(\frac{1}{f(x)}\) are both continuous (b) \(\tan (f(x))\) and \(\frac{1}{f(x)}\) are both discontinuous (c) \(\tan (f(x))\) and \(f^{\prime}(x)\) are both continuous (d) \(\tan (f(x))\) is continuous but \(\frac{1}{f(x)}\) is not
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