Problem 10
Question
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
Step-by-Step Solution
Verified Answer
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.
1Step 1: 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:
\(g'(x) = f'(x) - f'(1-x)\)
2Step 2: 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 \(0 < x < \frac{1}{2}\).
3Step 3: 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 \(\frac{1}{2} < x < 1\).
Key Concepts
Function DerivativeChain RuleIncreasing and Decreasing Functions
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.
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 \(-f'(1-x)\).
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.
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 \( g'(x) = f'(x) - f'(1-x) \).
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.
Other exercises in this chapter
Problem 7
Show that the function \(f(x)=\sin x, \quad x\) is a rationalno. \(=x, \quad x\) is an irrationalno. has positive derivative at \(x=0\) but \(f(x)\) is not incr
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Show that the function \(\begin{aligned} f(x) &=\frac{x}{2}+x^{2} \sin \frac{1}{x}, \quad x \neq 0 \\ &=0, \quad x=0 \end{aligned}\) is continuous and different
View solution Problem 11
Given \(g(x)=f\left(x^{2}-x-10\right)+f\left(14+x-x^{2}\right)\) and \(f^{\prime \prime}(x)>0\) for all real \(x\), except at finite no. of real values of \(x\)
View solution Problem 12
Investigate the following functions for extremum at \(x=0:-\) i. \(\quad f(x)=\sin x-x\). ii. \(\quad f(x)=\sin x-x+\frac{x^{3}}{3 !}\). iii. \(f(x)=\sin x-x+\f
View solution