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Chapter 1: Problem 93

$$ \text { If } f(x)=x+\frac{1}{x}, \text { prove that }(f(x))^{3}=f\left(x^{3}\right)+3 f\left(\frac{1}{x}\right) \text { . } $$

Short Answer

Expert verified
In summary, we proved the equation \((f(x))^3 = f(x^3) + 3f\left(\frac{1}{x}\right)\) for \(f(x) = x + \frac{1}{x}\) by finding expressions for \((f(x))^3\), \(f(x^3)\), and \(f\left(\frac{1}{x}\right)\), substituting them back into the given equation, and simplifying the expressions. The final result shows that both sides of the equation are equal, confirming the proof.

Step by step solution

01

Find the expression for \((f(x))^3\)

We are given that \(f(x) = x + \frac{1}{x}\). In the equation we want to prove, the first term is \((f(x))^3\). So, we need to cube the function. Next, we cube the function, \(f(x)\): \((f(x))^3 = \left(x + \frac{1}{x}\right)^3\)
02

Find the expression for \(f(x^3)\)

According to the given function, \(f(x) = x + \frac{1}{x}\). To find \(f(x^3)\), we need to replace \(x\) with \(x^3\): \(f(x^3) = x^3 + \frac{1}{(x^3)}\)
03

Find the expression for \(f\left(\frac{1}{x}\right)\)

Similarly, to find \(f\left(\frac{1}{x}\right)\), we need to replace \(x\) with \(\frac{1}{x}\) in the function \(f(x)\): \(f\left(\frac{1}{x}\right) = \frac{1}{x} + x\)
04

Substitute the expressions into the given equation

Now we substitute the expressions from Steps 1, 2, and 3 into the equation \((f(x))^3 = f(x^3) + 3f\left(\frac{1}{x}\right)\) to prove that the equation is true: \[\left(x + \frac{1}{x}\right)^3 = x^3 + \frac{1}{(x^3)} + 3\left(\frac{1}{x} + x\right)\]
05

Simplify and expand the terms

To verify that the above equation is true, we have to show that both sides of the equation simplify to the same expression. Expanding the left-hand side, we get: \((x + \frac{1}{x})^3 = x^3 + 3x^2\left(\frac{1}{x}\right) + 3x\left(\frac{1}{x}\right)^2 + \left(\frac{1}{x}\right)^3\) Simplifying this expression, we obtain: \(x^3 + 3x + \frac{3}{x} + \frac{1}{x^3}\) Now, simplifying the right-hand side: \(x^3 + \frac{1}{(x^3)} + 3\left(\frac{1}{x} + x\right) = x^3 + \frac{1}{x^3} + \frac{3}{x} + 3x\)
06

Compare both sides of the equation

Comparing the simplified expressions from Step 5, we find that: \(x^3 + 3x + \frac{3}{x} + \frac{1}{x^3} = x^3 + \frac{1}{x^3} + \frac{3}{x} + 3x\) Since both sides are equal, we have proven that \((f(x))^3 = f(x^3) + 3f\left(\frac{1}{x}\right)\), completing the exercise.

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Key Concepts

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

Differential Calculus
Differential Calculus is a branch of mathematics that studies the rates at which quantities change. It is the foundation for understanding motion, growth, and decay鈥攑roviding tools for determining the slope of a curve, optimization problems, and much more.

In the context of the IIT JEE syllabus, Differential Calculus covers critical concepts such as limits, continuity, differentiability, and the study of functions through their derivatives. The derivative of a function at a point gives the slope of the tangent to the curve at that point, which corresponds to the instantaneous rate of change of the function with respect to one of its variables.
Function Transformation
Function Transformation involves altering a function's formula to shift, reflect, stretch, or compress its graph. In our IIT JEE exercise, transformation is crucial in understanding how functions change when operations are applied to them.

Common transformations include scaling a function by a constant to stretch or compress it, translating a function by adding a constant to shift its graph vertically or horizontally, and reflecting a function about an axis. By mastering these transformations, students can deduce the behavior of complex functions from simple, base functions鈥攅nhancing their problem-solving toolkit for calculus.
Cubing a Function
Cubing a function means raising the function to the power of three. It is an operation that considerably changes the function's behavior, creating steeper curves and introducing potential inflection points. In our exercise, we cube the function f(x) = x + 1/x.

This process involves expanding the binomial to the third power, which can be done using the binomial theorem or by multiplying the function by itself three times. The result shows how the original simple rational function transforms into a more complex polynomial. Students should note that cubing a function emphasizes the function's behavior for large values of x and can lead to new local extremes and points of inflection.

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Most popular questions from this chapter

$$ \log _{2}\left(\frac{\sqrt[3]{4} \sqrt{2 \sqrt[5]{16}}}{\sqrt{2}}\right)-\log _{\frac{1}{2}} \sqrt[3]{\frac{4}{\sqrt{2}}}+\log _{\frac{1}{\sqrt{3}}}(9 \sqrt[3]{3}) $$

$$ 2^{3-\log _{4} 3}+7^{2 \log _{7} 2+1} $$

$$ \text { Given that } \log _{4} 125=a \text { , find } \log 64 \text { in terms of } a \text { . } $$

Find domain (write conditions only):- i. \(f(x)=\frac{1}{\ln (1-x)}+\sqrt{x+2}\) ii. \(f(x)=\sqrt{\sin ^{-1}\left(\log _{3} x\right)}\) iii. \(\quad f(x)=\sqrt{\ln (\sin x)}+\sin ^{-1}(\sqrt{\ln x})\) iv. \(\quad f(x)=\cos ^{-1}\left(\frac{3}{4+2 \sin x}\right)\) v. \(f(x)=\log _{2}\left(\log _{3} x\right)\) vi. \(\quad f(x)=\frac{x}{\ln \left(1+\sec ^{-1}(\ln x)\right)}\) vii. \(f(x)=\sqrt{4+x}-\sqrt{x+2}+\sqrt{15-x}\) viii. \(f(x)=\frac{1}{\sqrt{\ln \\{\cosh (\sin x)\\}}}\) ix. \(f(x)=\sqrt{-x}+\frac{1}{\sqrt{2+\operatorname{cosec}^{-1}(\sin x)}}\) x. \(\quad f(x)=2^{\frac{1}{\cos ^{-1} x}}+\cos ^{-1}\left(2^{x}\right)\) xi. \(\quad f(x)=\tan \left(\frac{1}{1-\tan ^{-1}\left(e^{x}\right)}\right)\) xii. \(f(x)=\sqrt[3]{\sin x}+\sqrt[4]{\cos x}\) xiii. \(f(x)=x^{x}\) xiv. \(f(x)=(\sin x)^{\cos ^{-1} x}\) xv. \(f(x)=\left(\frac{1+x}{1-x}\right)^{x}\) xvi. \(f(x)=\log _{\sin x} \cos x\)

$$ \sqrt{\log _{0.5}^{2} 4} $$
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