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

For a function to be continuous at \(x = 1\), \(\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{+}}f(x)=f(1)\). 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 \(x<1\). \(\lim_{x
ightarrow1^{-}}(5 - x)=4\).

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 \(1-\cos(x\pi)=2\sin^{2}\frac{x\pi}{2}\). \(\lim_{x
ightarrow1^{-}}\frac{1-\cos(x\pi)}{x + 1}=\frac{1-\cos(\pi)}{1 + 1}=\frac{1-(-1)}{2}=1\).

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 \(x<1\). \(\lim_{x
ightarrow1^{-}}(-x - 5)=-6\).

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

We know that \(\lim_{u
ightarrow0}\frac{\sin u}{u}=1\). Let \(u = 2x-2\), as \(x
ightarrow1\), \(u
ightarrow0\). \(\lim_{x
ightarrow1^{-}}\frac{\sin(2x - 2)}{x - 1}+1=\lim_{x
ightarrow1^{-}}\frac{\sin(2x - 2)}{x - 1}+1=2 + 1=3\).

Step6: Analyze the right - hand limit functions

For \(y=-x^{2}+4\) when \(x > 1\), \(\lim_{x
ightarrow1^{+}}(-x^{2}+4)=3\).
For \(y = 1\) when \(x>1\), \(\lim_{x
ightarrow1^{+}}1=1\).
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 \(\lim_{x
ightarrow1^{+}}\frac{1-\cos(4\pi x)}{2\pi^{2}(x - 1)^{2}}=\lim_{x
ightarrow1^{+}}\frac{2\sin^{2}(2\pi x)}{2\pi^{2}(x - 1)^{2}}\). 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 \(x
ightarrow1\), \(\lim_{x
ightarrow1^{+}}\frac{1-\cos(4\pi x)}{2\pi^{2}(x - 1)^{2}} = 8\).
For \(y=x^{3}-7\) when \(x>1\), \(\lim_{x
ightarrow1^{+}}(x^{3}-7)=-6\).

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