Problem 7
Question
In Exercises 7-26, find the numbers, if any, where the function is discontinuous. \(f(x)=2 x^{3}-3 x^{2}+4\)
Step-by-Step Solution
Verified Answer
The given function \(f(x) = 2x^3 - 3x^2 + 4\) is a polynomial function. Polynomial functions are continuous everywhere in their domain, so there are no discontinuities for this function.
1Step 1: Determine the function type
The given function is a polynomial function: \(f(x) = 2x^3 - 3x^2 + 4\).
Step 2: Analyze the continuity of polynomial functions
2Step 2: Continuity of polynomial functions
Polynomial functions are continuous everywhere in their domain, which means they do not have any discontinuities.
Step 3: State the result
3Step 3: Discontinuities in the given function
Since \(f(x)\) is a polynomial function, it is continuous everywhere. Thus, there are no values of x where this function is discontinuous.
Key Concepts
Continuity of FunctionsPolynomial FunctionsIdentifying Discontinuities
Continuity of Functions
When discussing the realm of calculus or basic algebra, one of the key concepts you'll encounter is the continuity of functions. A function is said to be continuous at a point if there are no interruptions, jumps, or breaks in its graph at that point. In more technical terms, a function is continuous at a point x = a if the following three conditions are met:
Now, let's connect this to a common misconception: for a function to be continuous, it doesn't have to be a straight line—it could have curves, bumps, or slopes, but what's important is that it doesn't abruptly stop or take a sudden detour.
- The function f(a) is defined (i.e., it has a value at a).
- The limit of the function as x approaches a exists.
- The limit of the function as x approaches a is equal to the function value at a, that is, \[\lim_{x \to a} f(x) = f(a)\].
Now, let's connect this to a common misconception: for a function to be continuous, it doesn't have to be a straight line—it could have curves, bumps, or slopes, but what's important is that it doesn't abruptly stop or take a sudden detour.
Polynomial Functions
Polynomial functions form a class of functions that are quite friendly, especially when considering their continuity. A polynomial function is made up of variables (or indeterminates), coefficients, and non-negative integer exponents. It will usually take the form \( f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_2x^2 + a_1x + a_0 \), where a_n, a_{n-1},...,a_0 are constants and a_n\( eq \)0. The largest exponent n is called the degree of the polynomial.
Now, why are we so fond of polynomial functions in the context of continuity? Simply put, they are continuous for all real numbers. That means, their graphs are unbroken lines or curves extending infinitely in both directions, with no holes, jumps, or vertical asymptotes to worry about. They are the 'smooth operators' of the function world, gliding from one x-value to the next without any unexpected interruptions.
Now, why are we so fond of polynomial functions in the context of continuity? Simply put, they are continuous for all real numbers. That means, their graphs are unbroken lines or curves extending infinitely in both directions, with no holes, jumps, or vertical asymptotes to worry about. They are the 'smooth operators' of the function world, gliding from one x-value to the next without any unexpected interruptions.
Identifying Discontinuities
Identifying discontinuities is a bit like being a detective in the world of mathematics. While polynomial functions don't have discontinuities, not all functions share this trait. Several types of discontinuities exist and recognizing them is crucial.
Firstly, there's the removable discontinuity, where the function has a hole at a point, which means the function value doesn't exist at that single point but is defined everywhere else close by. This often occurs in rational functions where the numerator and denominator share a common factor.
Next, we've got jump discontinuities. Picture this: you're graphing a function, and suddenly, it's like you've stepped off a cliff— the function jumps to a completely different value without passing through the intervening points.
Lastly, there's the infinite discontinuity, typically seen in cases with vertical asymptotes. The function's values get larger and larger (or smaller and smaller) approaching infinity as x gets closer to a certain point, but the function is never actually reached.
Being able to identify and classify these discontinuities is a fundamental skill in understanding function behaviors, particularly when you're analyzing more complex functions that aren't as straightforward as polynomials.
Firstly, there's the removable discontinuity, where the function has a hole at a point, which means the function value doesn't exist at that single point but is defined everywhere else close by. This often occurs in rational functions where the numerator and denominator share a common factor.
Next, we've got jump discontinuities. Picture this: you're graphing a function, and suddenly, it's like you've stepped off a cliff— the function jumps to a completely different value without passing through the intervening points.
Lastly, there's the infinite discontinuity, typically seen in cases with vertical asymptotes. The function's values get larger and larger (or smaller and smaller) approaching infinity as x gets closer to a certain point, but the function is never actually reached.
Being able to identify and classify these discontinuities is a fundamental skill in understanding function behaviors, particularly when you're analyzing more complex functions that aren't as straightforward as polynomials.
Other exercises in this chapter
Problem 5
Find the indicated limit. \(\lim _{x \rightarrow 1}\left(3 x^{2}-4 x+2\right)^{4}\)
View solution Problem 6
Find the indicated limit. \(\lim _{t \rightarrow 3}(2 t-1)^{2}\left(t^{2}-2 t\right)^{3}\)
View solution Problem 8
You are given \(\lim _{x \rightarrow a} f(x)=L\) and a tolerance \(\varepsilon\). Find a number \(\delta\) such that \(|f(x)-L|
View solution Problem 9
(a) use Equation (1) to find the slope of the secant line passing through the points \((a, f(a))\) and \((a+h, f(a+h)) ;\) (b) use the results of part (a) and E
View solution