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
\\(\frac{x^{2}-6x + 5}{|x - 1|}\\) when (x<1)
\\(\frac{1-cos(xpi)}{x + 1}\\) when (x<1)
\\(\frac{x^{2}+4x - 5}{|x - 1|}\\) when (x<1)
\\(\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 continuity condition

A function $y = f(x)$ 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 and match it with the right - hand limit as $x
ightarrow1^{+}$ for the functions $A$, $B$, $C$, $D$.

Step2: Analyze first function for $x<1$

For $y_1=\frac{x^{2}-6x + 5}{|x - 1|}$, factor the numerator: $x^{2}-6x + 5=(x - 1)(x - 5)$. When $x<1$, $|x - 1|=-(x - 1)$. So $y_1=\frac{(x - 1)(x - 5)}{-(x - 1)}=5 - x$. Then $\lim_{x
ightarrow1^{-}}y_1=4$. For $x>1$, for function $A:y=-x^{2}+4$, $\lim_{x
ightarrow1^{+}}(-x^{2}+4)=3$. So this is not a match.

Step3: Analyze second function for $x<1$

For $y_2=\frac{1-\cos(x\pi)}{x + 1}$, use the double - angle formula $1-\cos\theta=2\sin^{2}\frac{\theta}{2}$. Here $\theta=x\pi$, so $y_2=\frac{2\sin^{2}(\frac{x\pi}{2})}{x + 1}$. Then $\lim_{x
ightarrow1^{-}}\frac{2\sin^{2}(\frac{x\pi}{2})}{x + 1}=\frac{2\sin^{2}(\frac{\pi}{2})}{2}=1$. For $x>1$, for function $B:y = 1$, $\lim_{x
ightarrow1^{+}}1=1$.

Step4: Analyze third function for $x<1$

For $y_3=\frac{x^{2}+4x - 5}{|x - 1|}$, factor the numerator: $x^{2}+4x - 5=(x - 1)(x + 5)$. When $x<1$, $|x - 1|=-(x - 1)$. So $y_3=-(x + 5)$. Then $\lim_{x
ightarrow1^{-}}y_3=-6$. None of the $x>1$ functions match this limit.

Step5: Analyze fourth function for $x<1$

For $y_4=\frac{\sin(2x - 2)}{x - 1}+1$, use the fact that $\lim_{u
ightarrow0}\frac{\sin u}{u}=1$. Let $u = 2x-2$, as $x
ightarrow1$, $u
ightarrow0$. So $\lim_{x
ightarrow1^{-}}\frac{\sin(2x - 2)}{x - 1}+1=\lim_{x
ightarrow1^{-}}\frac{\sin(2x - 2)}{\frac{2x - 2}{2}}+1=2 + 1=3$. None of the $x>1$ functions match this limit.

Answer:

Second function: B, 1