Determine if Continuous f(x)=(x-7)/((x-1)(x+5))
The problem asks to determine whether the function f(x) = (x - 7) / ((x - 1)(x + 5)) is continuous. This involves checking the continuity of the function at all points in its domain. Continuity essentially means that there are no breaks, jumps, or holes in the graph of the function. To analyze this, typically, one would look at the behavior of the function as the variable x approaches any points where the denominator could be zero (since division by zero is undefined), and also ensure the function behaves well at other points. In this case, one would check the behavior around x = 1 and x = -5, because at these points the denominator of the function becomes zero, indicating potential points of discontinuity. However, a complete analysis must be done to formally conclude whether the function is continuous or not.
$f \left(\right. x \left.\right) = \frac{x - 7}{\left(\right. x - 1 \left.\right) \left(\right. x + 5 \left.\right)}$
Answer
Solution:
Step 1: Identify the domain to check for continuity of the function.
Step 1.1: Equate the denominator of $\frac{x - 7}{(x - 1)(x + 5)}$ to zero to find the points of discontinuity.
$(x - 1)(x + 5) = 0$
Step 1.2: Determine the values of $x$ that satisfy the equation.
Step 1.2.1: The product is zero if any of the factors is zero.
$x - 1 = 0$ $x + 5 = 0$
Step 1.2.2: Solve for $x$ when $x - 1$ is zero.
Step 1.2.2.1: Isolate $x$ by setting $x - 1$ to zero.
$x - 1 = 0$
Step 1.2.2.2: Solve for $x$ by adding $1$ to both sides.
$x = 1$
Step 1.2.3: Solve for $x$ when $x + 5$ is zero.
Step 1.2.3.1: Isolate $x$ by setting $x + 5$ to zero.
$x + 5 = 0$
Step 1.2.3.2: Solve for $x$ by subtracting $5$ from both sides.
$x = -5$
Step 1.2.4: Combine the solutions to find the values that make the denominator zero.
$x = 1, -5$
Step 1.3: Define the domain as all $x$ values that do not make the denominator zero.
Interval Notation: $(-\infty, -5) \cup (-5, 1) \cup (1, \infty)$ Set-Builder Notation: $\{x | x \neq 1, -5\}$
Step 2: The function $\frac{x - 7}{(x - 1)(x + 5)}$ is not continuous over the entire set of real numbers due to the points of discontinuity.
Discontinuous at $x = 1$ and $x = -5$
Knowledge Notes:
The problem involves determining the continuity of a rational function, which is a function that can be expressed as the ratio of two polynomials. The continuity of such a function depends on whether the denominator is zero at any point, as division by zero is undefined.
Key points to consider are:
Domain of a Function: The set of all possible input values (usually 'x') for which the function is defined. For rational functions, the domain excludes values that make the denominator zero.
Continuity: A function is continuous at a point if the limit of the function as it approaches the point from both sides is equal to the function's value at that point. A function is continuous over an interval if it is continuous at every point in that interval.
Discontinuity: Points where the function is not continuous. For rational functions, discontinuities occur where the denominator is zero.
Interval Notation: A way to represent intervals on the real number line. It uses parentheses to denote open intervals and brackets for closed intervals.
Set-Builder Notation: A notation for describing a set by stating the properties that its members must satisfy.
For the given function $f(x) = \frac{x - 7}{(x - 1)(x + 5)}$, the denominator is zero when $x = 1$ or $x = -5$. Therefore, the function is undefined and discontinuous at these points. The domain of the function is all real numbers except $x = 1$ and $x = -5$. The function is continuous for all other values of $x$ within its domain.
