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Chapter 21: Problem 28

Let \(f(x)=\frac{x}{\sqrt{a^{2}+x^{2}}}-\frac{d-x}{\sqrt{b^{2}+(d-x)^{2}}}, x \in \mathbf{R}\) where \(a, b\) and \(\mathrm{d}\) are non-zero real constants. Then : [Jan. 11, 2019 (II)] (a) \(f\) is an increasing function of \(x\) (b) \(f\) is a decreasing function of \(x\) (c) \(f^{\prime}\) is not a continuous function of \(x\) (d) \(f\) is neither increasing nor decreasing function of \(x\)

Short Answer

Expert verified
Function \( f(x) \) is neither increasing nor decreasing across its domain.

Step by step solution

01

Understand the Function

The given function is \( f(x) = \frac{x}{\sqrt{a^2 + x^2}} - \frac{d-x}{\sqrt{b^2 + (d-x)^2}} \). We need to determine the behavior of this function, specifically whether it is increasing, decreasing, or neither.
02

Calculate the Derivative

To analyze the behavior of \( f(x) \), first compute the derivative, \( f'(x) \). We'll use the quotient rule for the derivative of \( \frac{x}{\sqrt{a^2 + x^2}} \) and \( \frac{d-x}{\sqrt{b^2 + (d-x)^2}} \).
03

Derivative of First Term

Let \( u = x \) and \( v = \sqrt{a^2 + x^2} \), the derivative \( \frac{d}{dx} \frac{u}{v} \) using the quotient rule is \( \frac{v\cdot \frac{du}{dx} - u\cdot \frac{dv}{dx}}{v^2} \). For the first term, \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = \frac{x}{\sqrt{a^2 + x^2}} \), giving the derivative \( \frac{a^2}{(a^2 + x^2)^{3/2}} \).
04

Derivative of Second Term

For \( \frac{d-x}{\sqrt{b^2 + (d-x)^2}} \), let \( u = d-x \) and \( v = \sqrt{b^2 + (d-x)^2} \). Then \( \frac{du}{dx} = -1 \) and \( \frac{dv}{dx} = \frac{(d-x)}{\sqrt{b^2 + (d-x)^2}} \). Using the quotient rule, the derivative is \( \frac{-b^2}{(b^2 + (d-x)^2)^{3/2}} \).
05

Combine Derivatives

Combine the derivatives of the respective terms: \[ f'(x) = \frac{x}{(a^2 + x^2)^{3/2}} - \frac{b^2}{(b^2 + (d-x)^2)^{3/2}} \].
06

Analyze the Derivative

For \( f(x) \) to be increasing, \( f'(x) \) should be \( > 0 \) and for it to be decreasing, \( f'(x) \) should be \( < 0 \). Analyze each term: the derivative behaves differently on each side of its inputs, indicating that there is no unique behavior (either increasing or decreasing) across its domain.

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Key Concepts

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

Derivative
A derivative measures how a function changes as its input changes. It gives us the slope of the function at any given point along the curve.
This is useful in calculus because it allows us to determine where a function increases or decreases.
To find the derivative of a given function like the one in the exercise, we often use rules such as the quotient rule.
These help us calculate derivatives of more complex functions calculated as a fraction.
Increasing and Decreasing Functions
Understanding whether a function is increasing or decreasing helps us describe the behavior of the function across different values.
A function is increasing if its derivative is positive; this means as we increase the input, the output also increases.
  • If the derivative is consistently positive over an interval, the function is increasing on that interval.
  • Alternatively, if the derivative is negative, the function is decreasing over that interval.
However, in the case of the exercise's given function, the derivative switches signs, indicating the function has varying increases and decreases depending on the interval of interest.
Quotient Rule
The quotient rule is essential when we need to find the derivative of a function given as a ratio of two other functions.
For a function of the form \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), the derivative is given by: \[ \left( \frac{u}{v} \right)' = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]
In the exercise, we apply the quotient rule separately to two terms.
By doing so, we determine how each component of the function contributes to its overall slope.
Continuous Functions
A function is continuous if it has no jumps, breaks, or holes in its graph.
In terms of derivatives, if a function's derivative is continuous, it indicates smooth transitions without abrupt changes in direction.
The exercise hints at exploring whether the function's derivative itself is continuous.
Because the function \( f'(x) \) shows varying behaviors, particularly near critical points where the function changes from increasing to decreasing or vice versa, it suggests that \( f'(x) \) might not be continuous across its entire domain.

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Most popular questions from this chapter

If the tangents drawn to the hyperbola \(4 y^{2}=x^{2}+1\) intersect the co- ordinate axes at the distinct points \(A\) and \(B\), then the locus of the mid point of \(A B\) is[Online April 15, 2018] (a) \(x^{2}-4 y^{2}+16 x^{2} y^{2}=0\) (b) \(4 x^{2}-y^{2}+16 x^{2} y^{2}=0\) (c) \(4 x^{2}-y^{2}-16 x^{2} y^{2}=0\) (d) \(x^{2}-4 y^{2}-16 x^{2} y^{2}=0\)

A spherical iron ball of radius \(10 \mathrm{~cm}\) is coated with a layer of ice of uniform thickness that melts at a rate of \(50 \mathrm{~cm}^{3} / \mathrm{min}\). When the thickness of the ice is \(5 \mathrm{~cm}\), then the rate at which the thickness (in \(\mathrm{cm} / \mathrm{min}\) ) of the ice decreases, is : [April 10, 2019 (II)] (a) \(\frac{1}{18 \pi}\) (b) \(\frac{1}{36 \pi}\) (c) \(\frac{5}{6 \pi}\) (d) \(\frac{1}{9 \pi}\)

A point on the parabola \(y^{2}=18 x\) at which the ordinate increases at twice the rate of the abscissa is (a) \(\left(\frac{9}{8}, \frac{9}{2}\right)\) (b) \((2,-4)\) (c) \(\left(\frac{-9}{8}, \frac{9}{2}\right)\) (d) \((2,4)\)

If the function \(f\) given by \(f(x)=x^{3}-3(a-2) x^{2}+3 a x+7\), for some \(a \in R\) is increasing in \((0,1]\) and decreasing in \([1,5)\), then a root of the equation, \(\frac{f(x)-14}{(x-1)^{2}}=0(x \neq 1)\) is [Jan. 12, 2019 (II)] (a) \(-7\) (b) 5 (c) 7 (d) 6

Let \(m\) and \(M\) be respectively the minimum and maximum values of \(\quad\) [Sep. 06, 2020 (I)] \(\left|\begin{array}{ccc}\cos ^{2} x & 1+\sin ^{2} x & \sin 2 x \\ 1+\cos ^{2} x & \sin ^{2} x & \sin 2 x \\ \cos ^{2} x & \sin ^{2} x & 1+\sin 2 x\end{array}\right|\) Then the ordered pair \((m, M)\) is equal to : (a) \((-3,3)\) (b) \((-3,-1)\) (c) \((-4,-1)\) (d) \((1,3)\)
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