Determine if Continuous f(x)=( square root of x^2-25)/( natural log of x+9)
The problem involves examining the mathematical function f(x) = √(x^2 - 25) / (ln(x + 9)) to determine if it is continuous. Continuity of a function at a point means that the function is defined at that point, the limit of the function as it approaches the point exists, and the limit equals the function's value at that point. The question likely asks to investigate the behavior of the function across its entire domain, identify any points of discontinuity if they exist, and validate its continuity at points where it appears to be continuous by satisfying the definition of continuity.
$f \left(\right. x \left.\right) = \frac{\sqrt{x^{2} - 25}}{ln \left(\right. x + 9 \left.\right)}$
Answer
Solution:
Step:1
Identify the domain to assess continuity of the function.
Step:1.1
Ensure the argument inside $ln(x + 9)$ is positive to determine the valid domain. $x + 9 > 0$
Step:1.2
Isolate $x$ by subtracting $9$. $x > -9$
Step:1.3
The expression under the square root, $\sqrt{x^2 - 25}$, must be non-negative. $x^2 - 25 \geq 0$
Step:1.4
Find the values of $x$.
Step:1.4.1
Balance the inequality by adding $25$ to both sides. $x^2 \geq 25$
Step:1.4.2
Remove the square by taking the square root of both sides. $\sqrt{x^2} \geq \sqrt{25}$
Step:1.4.3
Simplify the resulting equation.
Step:1.4.3.1
Extract $x$ from under the radical sign. $|x| \geq \sqrt{25}$
Step:1.4.3.2
Simplify the square root of $25$.
Step:1.4.3.2.1
Express $25$ as $5^2$. $|x| \geq \sqrt{5^2}$
Step:1.4.3.2.1.2
Extract $5$ from under the radical. $|x| \geq |5|$
Step:1.4.3.2.1.3
Recognize that absolute value represents the distance from zero. $|x| \geq 5$
Step:1.4.4
Express $|x| \geq 5$ as two separate conditions.
Step:1.4.4.1
For the positive domain, consider $x \geq 0$.
Step:1.4.4.2
Without the absolute value, for positive $x$, we have $x \geq 5$.
Step:1.4.4.3
For the negative domain, consider $x < 0$.
Step:1.4.4.4
For negative $x$, remove the absolute value and multiply by $-1$. $-x \geq 5$
Step:1.4.4.5
Combine as a piecewise condition. $\begin{cases} x \geq 5 & \text{if } x \geq 0 \\ -x \geq 5 & \text{if } x < 0 \end{cases}$
Step:1.4.5
The intersection of $x \geq 5$ and $x \geq 0$ is $x \geq 5$.
Step:1.4.6
Invert the inequality $-x \geq 5$ by dividing by $-1$.
Step:1.4.6.1
Inverting the inequality gives $x \leq -5$.
Step:1.4.6.2
Simplify to find $x \leq -5$.
Step:1.4.7
Combine the solutions to get the domain. $x \leq -5$ or $x \geq 5$
Step:1.5
Set the denominator of $\frac{\sqrt{x^2 - 25}}{\ln(x + 9)}$ to zero to find undefined points. $\ln(x + 9) = 0$
Step:1.6
Solve for $x$.
Step:1.6.1
Use the properties of logarithms to rewrite the equation. $e^{\ln(x + 9)} = e^0$
Step:1.6.2
Convert the logarithmic equation to exponential form. $e^0 = x + 9$
Step:1.6.3
Isolate $x$.
Step:1.6.3.1
Rewrite as $x + 9 = e^0$.
Step:1.6.3.2
Since any number raised to zero is one, $x + 9 = 1$.
Step:1.6.3.3
Rearrange to solve for $x$.
Step:1.6.3.3.1
Subtract $9$ to get $x = 1 - 9$.
Step:1.6.3.3.2
Calculate to find $x = -8$.
Step:1.7
The domain is the set of $x$ values that make the function defined.
Interval Notation: $(-9, -8) \cup (-8, -5] \cup [5, \infty)$ Set-Builder Notation: $\{x | -9 < x \leq -5, x \geq 5, x \neq -8\}$
Step:2
The function $\frac{\sqrt{x^2 - 25}}{\ln(x + 9)}$ is not continuous across all real numbers due to the restricted domain.
Step:3
Knowledge Notes:
The problem involves determining the continuity of a function defined by $f(x) = \frac{\sqrt{x^2 - 25}}{\ln(x + 9)}$. To do this, we need to find the domain of the function, which is the set of all input values $x$ for which the function is defined. The domain is restricted by two conditions:
The argument of the natural logarithm, $\ln(x + 9)$, must be positive because the natural logarithm is only defined for positive numbers.
The expression under the square root, $\sqrt{x^2 - 25}$, must be non-negative because the square root is only defined for non-negative numbers.
After finding the domain, we check for any points where the function might not be continuous. This includes points where the denominator is zero since division by zero is undefined.
The solution involves several mathematical concepts:
Inequalities: Solving inequalities to find the domain of the function.
Absolute value: Understanding that $|x|$ represents the distance from zero, which leads to two separate cases depending on the sign of $x$.
Logarithms and exponentials: Using properties of logarithms and exponentials to solve equations involving the natural logarithm.
Continuity: Recognizing that a function is continuous if it is defined and does not have any breaks, jumps, or holes in its graph.
The final step is to express the domain in both interval notation and set-builder notation, which are two ways to describe sets of numbers. Interval notation uses intervals to describe the domain, while set-builder notation uses a condition that elements must satisfy to be included in the set.
