what are the open intervals of increase for the given rational function below? f(x)=1/(x^2 - 5x + 4) o (1…

Explanation:

Step1: Rewrite the denominator

First, factor the denominator $x^{2}-5x + 4=(x - 1)(x - 4)$. The function is $f(x)=\frac{1}{(x - 1)(x - 4)}=\frac{1}{x^{2}-5x + 4}$, and its domain is $x\neq1,x\neq4$.

Step2: Find the derivative

Use the quotient - rule. If $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 1$, $u^\prime=0$, $v=x^{2}-5x + 4$, $v^\prime=2x - 5$. So $f^\prime(x)=\frac{0\times(x^{2}-5x + 4)-1\times(2x - 5)}{(x^{2}-5x + 4)^{2}}=-\frac{2x - 5}{(x - 1)^{2}(x - 4)^{2}}$.

Step3: Determine where the derivative is positive

The function $f(x)$ is increasing when $f^\prime(x)>0$. Since $(x - 1)^{2}(x - 4)^{2}>0$ for all $x$ in the domain of $f(x)$ (the square of a non - zero real number is positive), we need to solve $-(2x - 5)>0$, which is equivalent to $2x-5<0$, so $x < 2.5$. Considering the domain $x\neq1,x\neq4$, the open intervals of increase are $(-\infty,1)\cup(1,2.5)$.

Answer:

$(-\infty,1)\cup(1,2.5)$