Question

In calculus, the critical numbers for a function are numbers in the domain of \(f\) where \(f^{\prime}(x)=0\) or \(f^{\prime}(x)\) is undefined. Find the critical numbers for \(f(x)=\frac{x^{2}-3 x+18}{x-2}\) if \(f^{\prime}(x)=\frac{x^{2}-4 x-12}{(x-2)^{2}}\)

Short answer

The critical numbers are \( x = 6 \) and \( x = -2 \).

Step-by-step solution

- Set the derivative equal to zero

To find the critical numbers, first set the derivative of the function to zero. Given: \[ f^{\text{'} }(x) = \frac{x^{2}-4 x-12}{(x-2)^{2}} \] Set \( f^{\text{'} }(x) = 0 \): \[ \frac{x^{2}-4 x-12}{(x-2)^{2}} = 0 \]

- Solve for x in the numerator

To solve the equation, focus on the numerator since a fraction is zero when its numerator is zero and the denominator is non-zero: \[ x^{2}-4 x-12 = 0 \] Factor the quadratic equation: \[ (x - 6)(x + 2) = 0 \]

- Find the x-values

Solve for the values of \(x\): \[ x - 6 = 0 \Rightarrow x = 6 \] \[ x + 2 = 0 \Rightarrow x = -2 \] These are the potential critical numbers.

- Check for undefined derivative

The derivative \( f^{\text{'} }(x) \) is undefined where the denominator equals zero: \[ (x-2)^{2} = 0 \Rightarrow x = 2 \] This implies that the derivative is undefined at \(x = 2\).

- Find the domain of the original function

Check the domain of the original function \( f(x) \). The function is undefined where the denominator equals zero: \[ x - 2 = 0 \Rightarrow x = 2 \] Thus, \( x = 6 \) and \( x = -2 \) are critical numbers within the domain of \( f(x) \).

Key concepts

derivative

A derivative represents how a function changes as its input changes. Simply put, the derivative of a function at a given point tells us the slope of the tangent line to the function's graph at that point. For a function \( f(x) \), its derivative is denoted as \( f^{\text{'} }(x) \) or \( \frac{d}{dx} f(x) \). The given function is:
\[ f(x) = \frac{x^{2} - 3x + 18}{x - 2} \]
and its derivative is:
\[ f^{\text{'} }(x) = \frac{x^{2} - 4x - 12}{(x - 2)^2} \]
In our step-by-step solution, we found the critical numbers by setting this derivative to zero and solving for \( x \). Remember, finding the critical numbers of a function involves determining where the derivative is either zero or undefined, as these points indicate possible maxima, minima or inflection points.

quadratic equation

A quadratic equation is a second-degree polynomial and can be written in the standard form \( ax^2 + bx + c = 0 \). To solve a quadratic equation, one can either factorize it, complete the square, or use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). In our problem, we derived a quadratic equation from the derivative's numerator:
\[ x^2 - 4x - 12 = 0 \]
We factorized this into two linear factors:
\[ (x - 6)(x + 2) = 0 \]
Then, we solved for \( x \) by setting each factor to zero:
\[ x - 6 = 0 \Rightarrow x = 6 \]
\[ x + 2 = 0 \Rightarrow x = -2 \]
This gives us the potential critical numbers.

domain of a function

The domain of a function refers to all possible input values (\( x \)) for which the function is defined. For rational functions, like our given \( f(x) = \frac{x^2 - 3x + 18}{x - 2} \), the function is undefined wherever the denominator is zero. To find the domain, we solve for when the denominator is zero:
\[ x - 2 = 0 \Rightarrow x = 2 \]
This tells us that \( x = 2 \) is not in the domain of \( f(x) \) since it would make the denominator zero, making the function undefined. Thus, the domain of \( f(x) \) is all real numbers except \( x = 2 \).

undefined function

A function is undefined at a point where it cannot produce a finite output. In our provided solution, we saw this with both the function \( f(x) \) and its derivative \( f^{\text{'} }(x) \). For this problem:
\[ f(x) = \frac{x^2 - 3x + 18}{x - 2} \]
The function is undefined at \( x = 2 \) because it would result in division by zero. Similarly, the derivative \( f^{\text{'} }(x) \) is:
\[ f^{\text{'} }(x) = \frac{x^2 - 4x - 12}{(x - 2)^2} \]
This is also undefined at \( x = 2 \), for the same reason. Identifying points where functions are undefined is crucial in understanding their behavior and ensuring correct solutions.