enter a letter and a number for each formula below so as to define a continuous function. the letter refers…

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 $\frac{x^{2}-6x + 5}{|x - 1|}$ 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 $x>1$

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 $\frac{x^{2}-6x + 5}{|x - 1|}$ 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 $x>1$

Answer

Explanation:

Step1: Recall continuity condition

For a function to be continuous at (x = 1), (\lim_{x\rightarrow1^{-}}f(x)=\lim_{x\rightarrow1^{+}}f(x)=f(1)). We will find the left - hand limit for each function when (x<1) and then match it with the right - hand limit of the candidate functions for (x > 1).

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

When (x<1), (|x - 1|=-(x - 1)). So (y=\frac{x^{2}-6x + 5}{|x - 1|}=\frac{(x - 1)(x - 5)}{-(x - 1)}=5 - x) for (x<1). (\lim_{x\rightarrow1^{-}}(5 - x)=4).

Step3: Analyze the function (\frac{1-\cos(x\pi)}{x + 1}) for (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}). (\lim_{x\rightarrow1^{-}}\frac{1-\cos(x\pi)}{x + 1}=\frac{1-\cos(\pi)}{1 + 1}=\frac{1-(-1)}{2}=1).

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

When (x<1), (|x - 1|=-(x - 1)) and (x^{2}+4x - 5=(x - 1)(x + 5)). So (y=\frac{x^{2}+4x - 5}{|x - 1|}=\frac{(x - 1)(x + 5)}{-(x - 1)}=-x - 5) for (x<1). (\lim_{x\rightarrow1^{-}}(-x - 5)=-6).

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

We know that (\lim_{u\rightarrow0}\frac{\sin u}{u}=1). Let (u = 2x-2), as (x\rightarrow1), (u\rightarrow0). (\lim_{x\rightarrow1^{-}}\frac{\sin(2x - 2)}{x - 1}+1=\lim_{x\rightarrow1^{-}}\frac{\sin(2x - 2)}{x - 1}+1=2 + 1=3).

Step6: Analyze the right - hand limit functions

For (y=-x^{2}+4) when (x > 1), (\lim_{x\rightarrow1^{+}}(-x^{2}+4)=3). For (y = 1) when (x>1), (\lim_{x\rightarrow1^{+}}1=1). For (y=\frac{1-\cos(4\pi x)}{2\pi^{2}(x - 1)^{2}}) when (x>1), using (1-\cos t = 2\sin^{2}\frac{t}{2}), let (t = 4\pi x), we have (\lim_{x\rightarrow1^{+}}\frac{1-\cos(4\pi x)}{2\pi^{2}(x - 1)^{2}}=\lim_{x\rightarrow1^{+}}\frac{2\sin^{2}(2\pi x)}{2\pi^{2}(x - 1)^{2}}). Using the fact that (\sin(2\pi x)=\sin(2\pi(x - 1)+2\pi)=\sin(2\pi(x - 1))\approx2\pi(x - 1)) as (x\rightarrow1), (\lim_{x\rightarrow1^{+}}\frac{1-\cos(4\pi x)}{2\pi^{2}(x - 1)^{2}} = 8). For (y=x^{3}-7) when (x>1), (\lim_{x\rightarrow1^{+}}(x^{3}-7)=-6).

The function (\frac{\sin(2x - 2)}{x - 1}+1) with (x<1) has a left - hand limit of 3, and the function (-x^{2}+4) with (x>1) has a right - hand limit of 3. The function (\frac{1-\cos(x\pi)}{x + 1}) with (x<1) has a left - hand limit of 1, and the function (1) with (x>1) has a right - hand limit of 1. The function (\frac{x^{2}+4x - 5}{|x - 1|}) with (x<1) has a left - hand limit of - 6, and the function (x^{3}-7) with (x>1) has a right - hand limit of - 6.

Answer:

a, 3; b, 1; D, - 6