# Consider the function f(x) = \frac{x^2 - 4x + 3}{(x-4)^2} a. Find the domain of the function....

## Question:

Consider the function {eq}f(x) = \frac{x^2 - 4x + 3}{(x-4)^2} {/eq}

a. Find the domain of the function.

b. Identify all critical points

c. Identify all inflection points

## Critical and Inflection Points:

Consider a function of one variable {eq}f(x). {/eq}

The critical points of the function are obtained setting the first derivative to zero, i.e. {eq}f'(x)=0. {/eq}

The inflection points of the function are found setting the second derivative to zero, i.e.

{eq}f''(x)=0. {/eq}

## Answer and Explanation:

Given the function

{eq}\displaystyle f(x) = \frac{x^2 - 4x + 3}{(x-4)^2} {/eq}

its first and second derivatives are calculated as

{eq}\displaystyle f'(x) =\frac{\frac{d}{dx}\left(x^2-4x+3\right)\left(x-4\right)^2-\frac{d}{dx}\left(\left(x-4\right)^2\right)\left(x^2-4x+3\right)}{\left(\left(x-4\right)^2\right)^2} \\ \displaystyle = \frac{\left(2x-4\right)\left(x-4\right)^2-2\left(x-4\right)\left(x^2-4x+3\right)}{\left(\left(x-4\right)^2\right)^2} \\ \displaystyle = \frac{2\left(-2x+5\right)}{\left(x-4\right)^3} \\ \displaystyle f''(x) =2\cdot \frac{\frac{d}{dx}\left(-2x+5\right)\left(x-4\right)^3-\frac{d}{dx}\left(\left(x-4\right)^3\right)\left(-2x+5\right)}{\left(\left(x-4\right)^3\right)^2} \\ \displaystyle = 2\cdot \frac{\left(-2\right)\left(x-4\right)^3-3\left(x-4\right)^2\left(-2x+5\right)}{\left(\left(x-4\right)^3\right)^2} \\ \displaystyle = -\frac{2\left(-4x+7\right)}{\left(x-4\right)^4}. {/eq}

a. Since the functions at numerator and denominator are continuous, the function f is everywhere continuous save at points

where the denominator is zero. Therefore the domain of the function is

{eq}\displaystyle (x-4)^2 \ne 0 \Rightarrow x \ne 4. {/eq}

b. The critical points of the function are found setting the first derivative to zero

{eq}\displaystyle f'(x) =0 \Rightarrow \frac{2\left(-2x+5\right)}{\left(x-4\right)^3} = 0 \\ \displaystyle \Rightarrow -2x+5 = 0 \\ \displaystyle \Rightarrow x = \frac{5}{2}. {/eq}

c. The inflection points are determined setting the second derivatives to zero

{eq}\displaystyle f''(x) =0 \Rightarrow -\frac{2\left(-4x+7\right)}{\left(x-4\right)^4} = 0 \\ \displaystyle \Rightarrow -4x+7 = 0 \\ \displaystyle \Rightarrow x = \frac{7}{4}. {/eq}

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from Math 104: Calculus

Chapter 10 / Lesson 6