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Chapter 5: Problem 10

If \(g(x)=f(x)+f(1-x)\) and \(f^{\prime \prime}(x)<0 ; 0 \leq x \leq 1\), show that \(g(x)\) increases in \(0

Short Answer

Expert verified
In the interval \(0 < x < \frac{1}{2}\), \(g'(x) = f'(x) - f'(1-x) < 0\), so \(g(x)\) is decreasing. In the interval \(\frac{1}{2} < x < 1\), \(g'(x) = f'(x) - f'(1-x) > 0\), so \(g(x)\) is increasing.

Step by step solution

01

Find the derivative of g(x)

First, we will find the derivative (first derivative) of \(g(x)\). Apply the chain rule to differentiate \(f(1-x)\). The chain rule states that the derivative of a composite function is the derivative of the outer function multiplied by the derivative of the inner function: \(g'(x) = f'(x) - f'(1-x)\)
02

Determine the sign of g'(x) in 0 < x < 0.5

In order to determine whether \(g(x)\) is increasing or decreasing in the interval \(0 < x < \frac{1}{2}\), we must analyze the sign of its derivative, \(g'(x)\). Recall that if a function's derivative is positive in a given interval, the function is increasing in that interval. If the derivative is negative, the function is decreasing. In the given interval \(0 < x < \frac{1}{2}\), we can see: \(g'(x) = f'(x) - f'(1-x)\) Since \(f''(x) < 0\), we have \(f'(1-x) > f'(x)\). It follows that in this interval, \(g'(x) < 0\). Thus, \(g(x)\) is decreasing in the interval \(0 < x < \frac{1}{2}\).
03

Determine the sign of g'(x) in 0.5 < x < 1

Now we will find the behavior of \(g(x)\) in the interval \(\frac{1}{2} < x < 1\). Similarly to step 2, we analyze the sign of \(g'(x)\): \(g'(x) = f'(x) - f'(1-x)\) In the given interval \(\frac{1}{2} < x < 1\), we have: Since \(f''(x) < 0\), we have \(f'(x) > f'(1-x)\). It follows that in this interval, \(g'(x) > 0\). Thus, \(g(x)\) is increasing in the interval \(\frac{1}{2} < x < 1\).

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Key Concepts

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

Function Derivative
The derivative of a function gives us a powerful tool to understand its behavior. In simple terms, the derivative represents the rate at which the function changes as its input changes. This rate of change is one of the cornerstones of differential calculus and helps us understand how functions behave at different points.

To find the derivative of a function like \( g(x) = f(x) + f(1-x) \), we look at each part separately.
  • The derivative of \( f(x) \) is \( f'(x) \).
  • For \( f(1-x) \), we use the chain rule to find its derivative as \(-f'(1-x)\).
Putting it together, \( g'(x) = f'(x) - f'(1-x) \). This equation gives us the derivative of the entire function \( g(x) \) and provides insight into how \( g(x) \) changes across different intervals.
Chain Rule
The chain rule is a method used to differentiate composite functions. A composite function is formed when one function is nested inside another, like \( f(1-x) \) in our example. The chain rule helps us find the derivative of such functions by considering both the "outer function" and the "inner function."

In our case, the outer function is \( f(u) \) where \( u = 1-x \). The inner function is \( 1-x \) itself. The derivative of the outer function \( f(u) \) is \( f'(u) \) and for the inner function \( 1-x \), the derivative is \(-1\). Multiplying these using the chain rule gives \( -f'(1-x) \), showing how each part contributes to the change in \( f(1-x) \) due to a change in \( x \).

Using the chain rule simplifies problems involving nested functions and helps us easily compute derivatives that would be more complex otherwise.
Increasing and Decreasing Functions
Understanding whether a function is increasing or decreasing is crucial in determining its overall behavior. The derivative of a function lets us examine this behavior over specific intervals.

When the derivative of a function \( g(x) \) is positive over an interval, \( g(x) \) is increasing over that interval. Conversely, if the derivative is negative, the function is decreasing.

In our specific exercise, we assess \( g(x) = f(x) + f(1-x) \) using its derivative \( g'(x) = f'(x) - f'(1-x) \).
  • For \( 0 < x < \frac{1}{2} \), since \( f'(1-x) > f'(x) \) (because \( f''(x) < 0 \)), \( g'(x) < 0 \), indicating that \( g(x) \) decreases in this range.
  • In contrast, for \( \frac{1}{2} < x < 1 \), \( f'(x) > f'(1-x) \), making \( g'(x) > 0 \), and thus \( g(x) \) increases over this interval.
These insights are invaluable when sketching graphs or looking for trends in functions.

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Most popular questions from this chapter

For what value of \(a\), the mean rate of change of the function \(f(x)=x^{3}\) in the interval \([-1, a]\) is equal to the instantaneous rate of change at \(a ?\)

If \(2 a+3 b+6 c=0\), then show that the equation \(a x^{2}+b x+c=0\) has at least one real root between 0 and \(1 .\)

Test the indicated points for point of inflection:- i. \(f(x)=x^{3}-5 x^{2}+3 x-5\) at \(x=1, \frac{5}{3}, 2\). ii. \(f(x)=x^{4}-12 x^{3}+48 x^{2}\) at \(x=1,2,3,4\). iii. \(f(x)=x+x^{\frac{5}{3}}-2\) at \(x=0\). iv. \(f(x)=x^{2}+x^{\frac{4}{3}}+1\) at \(x=0\). v. \(f(x)=x+x^{\frac{2}{3}}+4\) at \(x=0\). vi. \(f(x)=x+x^{\frac{3}{5}}-3\) at \(x=0\). vii. \(f(x)=\sin x+\frac{x^{3}}{3 !}-\frac{x^{5}}{5 !}\) at \(x=0\). viii. \(f(x)=e^{x}-\frac{x^{2}}{2}-\frac{x^{3}}{6}\) at \(x=0\). ix. \(\quad f(x)=\sin x, \quad x \geq 0\) \(=x-\frac{x^{3}}{6}, \quad x<0\) at \(x=0\). \\{Ans. 0 is point of inflection\\}

Prove that:- i. \(\quad e^{x}>1+x, \quad x \neq 0\). ii. \(\quad x-\frac{x^{3}}{6}<\sin x0\). iii. \(\frac{x}{1+x} \leq \ln (1+x) \leq x, \quad x>-1\). iv. \(\frac{x}{1+x^{2}}<\tan ^{-1} x0\). v. \(\quad \ln x>\frac{2(x-1)}{x+1}, \quad x>1\). vi. \(\quad 2 x \tan ^{-1} x \geq \ln \left(1+x^{2}\right)\). vii. \(\ln (1+x)>\frac{\tan ^{-1} x}{1+x}, \quad x>0\). viii. \(\sin x0\). ix. \(\quad \sin x+\tan x>2 x, \quad 01+\frac{x^{2}}{2}, \quad x \neq 0\). xi. \(1+x \ln \left(x+\sqrt{x^{2}+1}\right) \geq \sqrt{1+x^{2}}\). xii. \(\quad 2 \sin x+\tan x \geq 3 x\) for \(0 \leq x<\frac{\pi}{2}\).

The function \(f(x)=a \sin x+\frac{1}{3} \sin 3 x\) has maximum value at \(x=\frac{\pi}{3}\). Find the value of \(a\).
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