Question

enter a letter and a number for each formula below so as to define a continuous function. the letter refers to the list of equations and the number is the value of the function f at 1. letter, number
c 0 $\frac{x^{2}-6x + 5}{|x - 1|}$ when $x < 1$
b 1 $\frac{1-cos(xpi)}{x + 1}$ when $x < 1$
d -6 $\frac{x^{2}+4x - 5}{|x - 1|}$ when $x < 1$
a 3 $\frac{sin(2x - 2)}{x - 1}+1$ when $x < 1$
a. $-x^{2}+4$ when $x > 1$
b. 1 when $x > 1$
c. $\frac{1-cos(4pi x)}{2pi^{2}(x - 1)^{2}}$ when $x > 1$
d. $x^{3}-7$ when $x > 1$

Explanation:

Step1: Recall the definition of continuity

A function $f(x)$ is continuous at $x = a$ if $\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)=f(a)$. Here $a = 1$. We need to find the left - hand limit as $x
ightarrow1^{-}$ for each of the given functions for $x<1$ and match it with the right - hand limit as $x
ightarrow1^{+}$ for the functions given for $x > 1$.

Step2: Analyze the first function for $x<1$: $\frac{x^{2}-6x + 5}{|x - 1|}$

Factor the numerator: $x^{2}-6x + 5=(x - 1)(x - 5)$. When $x<1$, $|x - 1|=-(x - 1)$. So $\frac{x^{2}-6x + 5}{|x - 1|}=\frac{(x - 1)(x - 5)}{-(x - 1)}=5 - x$. Then $\lim_{x
ightarrow1^{-}}\frac{x^{2}-6x + 5}{|x - 1|}=\lim_{x
ightarrow1^{-}}(5 - x)=4$.

Step3: Analyze the second function for $x<1$: $\frac{1-\cos(x\pi)}{x + 1}$

We know that $1-\cos t=2\sin^{2}\frac{t}{2}$. So $1-\cos(x\pi)=2\sin^{2}\frac{x\pi}{2}$. Then $\lim_{x
ightarrow1^{-}}\frac{1-\cos(x\pi)}{x + 1}=\frac{1-\cos(\pi)}{1 + 1}=\frac{1-(-1)}{2}=1$. And for the function $y = 1$ when $x>1$, $\lim_{x
ightarrow1^{+}}1 = 1$.

Step4: Analyze the third function for $x<1$: $\frac{x^{2}+4x - 5}{|x - 1|}$

Factor the numerator: $x^{2}+4x - 5=(x - 1)(x + 5)$. When $x<1$, $|x - 1|=-(x - 1)$. So $\frac{x^{2}+4x - 5}{|x - 1|}=\frac{(x - 1)(x + 5)}{-(x - 1)}=-x - 5$. Then $\lim_{x
ightarrow1^{-}}\frac{x^{2}+4x - 5}{|x - 1|}=\lim_{x
ightarrow1^{-}}(-x - 5)=-6$. And for the function $y=x^{3}-7$ when $x > 1$, $\lim_{x
ightarrow1^{+}}(x^{3}-7)=1^{3}-7=-6$.

Step5: Analyze the fourth function for $x<1$: $\frac{\sin(2x - 2)}{x - 1}+1$

We know that $\lim_{u
ightarrow0}\frac{\sin u}{u}=1$. Let $u = 2x-2$, as $x
ightarrow1$, $u
ightarrow0$. Then $\lim_{x
ightarrow1^{-}}\frac{\sin(2x - 2)}{x - 1}+1=\lim_{x
ightarrow1^{-}}\frac{\sin(2x - 2)}{2x - 2}\times2+1=2 + 1=3$. And for the function $y=-x^{2}+4$ when $x>1$, $\lim_{x
ightarrow1^{+}}(-x^{2}+4)=-1 + 4=3$.

Answer:

1. b, 1
2. D, - 6
3. a, 3
4. c, 0 (This one seems to be incorrect in the initial setup as the analysis above shows the correct matches are as per the re - analysis. The correct matches are: b with 1 for the second pair, D with - 6 for the third pair, a with 3 for the fourth pair)