Problem 31 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(f(x)=-1+|x-1|, \quad-1 \leq x \leq 3\) \(g(x)=2-|x+1|, \quad-2 \leq x \leq 2\), then calculate \((\) fog \() x\) and \((\) gof \() x\). Draw their graphs. Discuss the continuity of \((\) fog \() x\) at \(x=1\) and $$ \text { differentiability of }(\text { gof }) x \text { at } x=1 \text { . } $$

Step-by-Step Solution

Answer

The function \(f(g(x))\) can be expressed as a piecewise function: \(f(g(x))= \[ \begin{cases} -1 + |2+x| & \text{if } x < -1 \\ -1 + |-x| & \text{if } x \geq -1 \end{cases} \] \) And the function \(g(f(x))\) can be expressed as a piecewise function: \(g(f(x))= \[ \begin{cases} 2 - |1-x| & \text{if } x < 1 \\ 2 - |x| & \text{if } x \geq 1 \end{cases} \] \) The function f(g(x)) is continuous at x = -1 and the function g(f(x)) is not differentiable at x = 1.

Step 1: Express g(x) in piecewise form

First, express the function g(x) as a piecewise function. This will make calculating f(g(x)) and g(f(x)) easier. g(x) can be expressed as: g(x) = \[ \begin{cases} 2 - (-(x+1)) & \text{if } x < -1 \\ 2 - (x+1) & \text{if } x \geq -1 \end{cases} \] Now we can move on to calculate f(g(x)) and g(f(x)).

Step 2: Calculate the f(g(x))

Plug g(x) into the function of f(x): \(f(g(x)) = -1 + |g(x) - 1|\) Now consider the two cases of g(x): 1. If \(x<-1\): \(f(g(x)) = -1 + |(2 - (-(x + 1))) - 1| = -1 + |3+x- 1| = -1+|2+x|\) 2. If \(x\geq-1\): \(f(g(x)) = -1 + |(2 - (x + 1)) - 1| = -1 + |1-x-1| = -1 + |-x|\) Now f(g(x)) can be written in piecewise form: \(f(g(x))= \[ \begin{cases} -1 + |2+x| & \text{if } x < -1 \\ -1 + |-x| & \text{if } x \geq -1 \end{cases} \] \)

Step 3: Calculate g(f(x))

Plug f(x) into g(x): \(g(f(x)) = 2 - |f(x) + 1|\) In this case, since g(x) has a modulus function, rewrite f(x) in piecewise form: f(x) = \[ \begin{cases} -1 + (-(x-1)) & \text{if } x < 1 \\ -1 + (x-1) & \text{if } x \geq 1 \end{cases} \] Now consider the two cases of f(x): 1. If \(x<1\): \(g(f(x)) = 2 - |(-1 + (-(x - 1)) + 1)| = 2 - |1 - x|\) 2. If \(x \geq 1\): \(g(f(x))= 2 - |(-1 + (x - 1) + 1)| = 2 - |x|\) Now g(f(x)) can be written in piecewise form: \(g(f(x))= \[ \begin{cases} 2 - |1-x| & \text{if } x < 1 \\ 2 - |x| & \text{if } x \geq 1 \end{cases} \] \) #Step 2: Draw the graphs#

Step 4: Draw graphs

Using any graphing software or tool, draw the graphs of both f(g(x)) and g(f(x)) with respect to their piecewise defined functions. The following are important key points: 1. For f(g(x)): - Check the endpoints x = -2 and x = 2. - Check the transition at x = -1. 2. For g(f(x)): - Check the endpoints x = -1 and x = 3. - Check the transition at x = 1. #Step 3: Analyze continuity of f(g(x)) at x=1#

Step 5: Analyze continuity

We have the following piecewise function for f(g(x)): \(f(g(x))= \[ \begin{cases} -1 + |2+x| & \text{if } x < -1 \\ -1 + |-x| & \text{if } x \geq -1 \end{cases} \] \) Let's check the values at x=-1: \(f(g(-1)) = -1 +|-(-1)| = 0\) Since the function in both cases (x<-1 and x≥-1) gives the same value at x = -1, f(g(x)) is continuous at x = -1. #Step 4: Analyze differentiability of g(f(x)) at x=1#

Step 6: Analyze differentiability

We have the following piecewise function for g(f(x)): \(g(f(x))= \[ \begin{cases} 2 - |1-x| & \text{if } x < 1 \\ 2 - |x| & \text{if } x \geq 1 \end{cases} \] \) Now, we need to check the derivatives for both cases at x=1 for the function g(f(x)): If \(x<1\), \(g(f(x))=2-(1-x)=1+x\), so its derivative is \(g'(f(x))=1\). If \(x\geq 1\), \(g(f(x))= 2-|x| = 2-x\), so its derivative is \(g'(f(x))=-1\). Since the left and right derivatives at x=1 are different, it means the function g(f(x)) is not differentiable at x=1.

Key Concepts

Understanding Piecewise Functions

Piecewise functions are multifaceted; they can have different definitions over different intervals. Imagine them as a chameleon shifting appearances depending on the environment. In the realm of functions, their 'environment' is determined by the input value, or 'x'.

For instance, in the original exercise, the functions are defined in intervals that pivot around the points -1 and 1. This choice of intervals is not random but dictates where each function rule applies. When dealing with piecewise functions, we approach each interval separately and apply the corresponding function rule to calculate the output for that segment of the domain.

• For values of 'x' less than -1, apply one function rule.
• For values of 'x' greater than or equal to -1, switch to the other rule.

This segregation allows us to describe complex behavior, such as abrupt changes in direction or gaps that a single function rule couldn't account for comprehensively.

Graphing Piecewise Functions Skillfully

To graph a piecewise function, think of it as tracing a mountain landscape, where each section has its own unique slope and terrain.

Begin with a simple sketch of the x-y axis, and mark the critical points where the function changes its rule. From there, draw each segment in its respective interval, bearing in mind to:

• Consider the domain restrictions (don't go beyond your defined 'x' values).
• Ensure you understand the behavior at the boundaries—is the function approaching an endpoint, leaping over it, or maybe coming to a halt right there?

Accuracy is key, and don't forget to check both endpoints of each interval, as they can provide valuable clues about how the graph transitions from one section to another. By combining the segments carefully, you'll watch the piecewise function come to life on your graph.

The Seamless Story of Continuity

Continuity is the mathematical equivalent of a smooth melody, without any jarring notes or pauses. A function is considered continuous at a point if you can sketch its graph at that point without lifting your pencil.

In more precise terms, for a function to be continuous at a point 'x=c', three conditions must be met:

• The function must be defined at 'x=c'.
• The limit as 'x' approaches 'c' from the left and the right must exist.
• The limit as 'x' approaches 'c' and the function's value at 'c' must be the same.

For f(g(x)) in our exercise, continuity is confirmed at x = -1 because it passes these tests with flying colors: the values on either side of x = -1 match, and there's no disruption as you ease your way along the curve.

Differentiability: The Smooth Turn Check

A function's differentiability is a bit like a car's ability to take a smooth turn. If a function is differentiable at a point, it means there's a clear, defined direction at that very point—you can take that 'turn' without any sudden jerks.

Mathematically, a function is differentiable at 'x=c' if its derivative exists at that point. This involves having a single, unambiguous tangent line, implying the function can't have any sharp kinks or corners there. To assess this, you check the slope of the function as 'x' approaches 'c' from both directions. If the slopes (derivatives) match up nicely, you've got differentiability.

However, for g(f(x)) at x = 1, we notice a breakdown in this smooth transition. The derivative leaps from 1 to -1, like a car hitting a curb; hence, we conclude that g(f(x)) is not differentiable at x = 1. It’s the point where the function's graph would make a sharp turn, disrupting the otherwise smooth journey of the graphing pencil.