Problem 48 - Calculus: Graphical, Numerical, Algebraic

Question

Continuous Function Find a value for \(a\) so that the function \(f(x)=\left\\{\begin{array}{ll}{2 x+3,} & {x \leq 2} \\ {a x+1,} & {x>2}\end{array}\right.\) is continuous.

Step-by-Step Solution

Answer

The value for \(a\) that makes the function continuous is 3.

Step 1: Identify the transition point

The transition point is given by the condition in the function definition, which is \(x = 2\). This is where the two pieces of the function meet.

Step 2: Evaluate the first part of the function at the transition point

Plug \(x = 2\) into the first part of the function, which gives: \(2*2 + 3 = 7\).

Step 3: Evaluate the second part of the function at the transition point

Now plug \(x = 2\) into the second part of the function, which should equal the value from step 2 to ensure continuity: \(a*2 + 1 = 7\).

Step 4: Solve for \(a\)

Solve the equation for \(a\): \(a = \frac{(7 -1)}{2}\).

Key Concepts

Piecewise Functions

When studying functions in mathematics, one may encounter a special type referred to as a piecewise function. These functions are defined by multiple sub-functions, each applied to a certain interval of the domain. In essence, a piecewise function is a combination of different expressions, each governing its own part of the function's domain.

This construct allows for greater flexibility, as each 'piece' of the function can represent a different behavior or rule. For example, you might have a different formula for calculating taxes below and above a certain income threshold, effectively making the tax calculation a piecewise function.

To ensure a continuous piecewise function, the output from each sub-function must match at the borders where the function 'pieces' intersect. In our exercise, this concept is illustrated by requiring the two expressions defined for portions of the domain to yield the same function value at the point where they meet, which is at the transition point, here, at x = 2.

Limits in Calculus

The concept of limits is fundamental in calculus, and it plays a pivotal role in defining the notion of continuity for functions. A limit describes the value that a function approaches as the input (or x-value) approaches a certain point.

When dealing with piecewise functions, limits from the left and limits from the right at a given point must be the same for the function to be continuous at that point. To determine if the given function f(x) is continuous at the transition point x = 2, we must ensure that the limit of f(x) as x approaches 2 from both sides is the same. This is known as having matching one-sided limits at the point of interest.

In our exercise, the limit of 2x + 3 as x approaches 2 is the same as the value obtained by directly substituting 2 into the expression, yielding 7. For the other piece, ax + 1, we must ensure that its limit as x approaches 2 also equals 7 to maintain continuity, thus determining the correct value of a.

Solving for Variables

In algebra and calculus alike, an essential skill is solving for an unknown variable. The process typically involves isolating the variable on one side of the equation to determine its value. To effectively solve for a variable, one must apply inverse operations such as subtraction, division, and factoring, while adhering to the properties of equality.

In the context of our exercise, solving for the variable a in the equation 2a + 1 = 7 serves to find the value that ensures function continuity. This is done by first subtracting 1 from both sides, obtaining 2a = 6, followed by dividing both sides by 2, yielding a = 3.

This straightforward approach illustrates finding a particular value of a variable that is crucial to satisfying a greater condition—in this case, the continuity of a piecewise function. Understanding how to manipulate and solve for variables is a key component of mastering higher-level mathematics.