Question

let (f(x)) be defined as follows. (f(x)=\frac{6x - 6}{x^{4}-12x^{3}+36x^{2}}) find each (x) - value of discontinuity of (f(x)). for each, give the (x) - value of the point of discontinuity and evaluate the one - sided limits. if you have more than one (x) - value, give them in numerical order, from smallest to largest. enter dne in any extra answer boxes. a. value: (x = a=) (lim_{x
ightarrow a^{-}}f(x)=) (lim_{x
ightarrow a^{+}}f(x)=) b. value: (x = b=) (lim_{x
ightarrow b^{-}}f(x)=) (lim_{x
ightarrow b^{+}}f(x)=) c. value: (x = c=) (lim_{x
ightarrow c^{-}}f(x)=)

Explanation:

Step1: Find points of discontinuity

Discontinuities occur where the denominator $x^{4}-12x^{3}+36x^{2}=x^{2}(x - 6)^{2}=0$. Solving $x^{2}(x - 6)^{2}=0$ gives $x = 0$ and $x=6$.

Step2: Simplify the function

Factor the numerator and denominator: $f(x)=\frac{6(x - 1)}{x^{2}(x - 6)^{2}}$.

Step3: Calculate left - hand limit as $x\to0$

$\lim_{x\to0^{-}}\frac{6(x - 1)}{x^{2}(x - 6)^{2}}$. As $x\to0^{-}$, the numerator approaches $6\times(- 1)=-6$ and the denominator is a positive number approaching 0. So, $\lim_{x\to0^{-}}f(x)=-\infty$.

Step4: Calculate right - hand limit as $x\to0$

$\lim_{x\to0^{+}}\frac{6(x - 1)}{x^{2}(x - 6)^{2}}$. As $x\to0^{+}$, the numerator approaches $-6$ and the denominator is a positive number approaching 0. So, $\lim_{x\to0^{+}}f(x)=-\infty$.

Step5: Calculate left - hand limit as $x\to6$

$\lim_{x\to6^{-}}\frac{6(x - 1)}{x^{2}(x - 6)^{2}}$. Let $t=x - 6$, then $x=t + 6$. The function becomes $\frac{6(t + 5)}{(t + 6)^{2}t^{2}}$. As $t\to0^{-}$, the numerator approaches $6\times5 = 30$ and the denominator is a positive number approaching 0. So, $\lim_{x\to6^{-}}f(x)=+\infty$.

Step6: Calculate right - hand limit as $x\to6$

$\lim_{x\to6^{+}}\frac{6(x - 1)}{x^{2}(x - 6)^{2}}$. Let $t=x - 6$, then $x=t + 6$. The function becomes $\frac{6(t + 5)}{(t + 6)^{2}t^{2}}$. As $t\to0^{+}$, the numerator approaches $30$ and the denominator is a positive number approaching 0. So, $\lim_{x\to6^{+}}f(x)=+\infty$.

Answer:

a. Value: $x = 0$
$\lim_{x\to0^{-}}f(x)=-\infty$
$\lim_{x\to0^{+}}f(x)=-\infty$
b. Value: $x = 6$
$\lim_{x\to6^{-}}f(x)=+\infty$
$\lim_{x\to6^{+}}f(x)=+\infty$
c. Value: $x =$ DNE
$\lim_{x\to c^{-}}f(x)=$ DNE
$\lim_{x\to c^{+}}f(x)=$ DNE