Answer:
See below
Step-by-step explanation:
Problem 1
Since [tex]ax-b[/tex] (left-side) and [tex]2x^2+3ax+b[/tex] (right-side) both approach [tex]-1[/tex], then we need to set them equal to each other in order for the limit to exist:
[tex]\lim_{x \to -1^-} f(x)= \lim_{x \to -1^+} f(x)[/tex]
[tex]ax-b=2x^2+3ax+b[/tex]
[tex]a(-1)-b=2(-1)^2+3a(-1)+b[/tex]
[tex]-a-b=2-3a+b[/tex]
[tex]2a-b=2+b[/tex]
[tex]2a-2b=2[/tex]
Since [tex]2x^2+3ax+b[/tex] (left-side) and [tex]4[/tex] (right-side) both approach [tex]1[/tex], they need to be set equal to each other in order for the limit to exist:
[tex]\lim_{x \to 1^-} f(x)= \lim_{x \to 1^+} f(x)[/tex]
[tex]2x^2+3ax+b=4[/tex]
[tex]2(1)^2+3a(1)+b=4[/tex]
[tex]2+3a+b=4[/tex]
[tex]3a+b=2[/tex]
Now simply solve the system of equations:
[tex]\left \{ {{2a-2b=2} \atop {3a+b=2}} \right.[/tex]
[tex]\left \{ {{2a-2b=2} \atop {6a+2b=4}} \right.[/tex]
[tex]8a=6[/tex]
[tex]a=\frac{6}{8}[/tex]
[tex]a=\frac{3}{4}[/tex]
[tex]3a+b=2[/tex]
[tex]3(\frac{3}{4})+b=2[/tex]
[tex]\frac{9}{4}+b=2[/tex]
[tex]b=-\frac{1}{4}[/tex]
Therefore, the piecewise function is continuous for all the values of x when [tex]a=\frac{3}{4}[/tex] and [tex]b=-\frac{1}{4}[/tex].
Problem 2:
Vertical Asymptotes:
There are no vertical asymptotes since the denominator can never be 0 with real numbers.
Horizontal Asymptotes:
We define a horizontal asymptote of a function as the limit as x approaches infinity (or negative infinity):
[tex]f(x)=\frac{3x-2}{\sqrt{4x^2+5}}[/tex]
[tex]\lim_{x \to \infty}=\frac{3x-2}{\sqrt{4x^2+5}}[/tex]
[tex]\lim_{x \to \infty}=\frac{3-\frac{2}{x}}{\frac{1}{x} \sqrt{4x^2+5}}[/tex]
[tex]\lim_{x \to \infty}=\frac{3-\frac{2}{x}}{ \sqrt{\frac{1}{x^2}(4x^2+5)}}[/tex]
[tex]\lim_{x \to \infty}=\frac{3-\frac{2}{x}}{ \sqrt{4+\frac{5}{x^2}}}[/tex]
[tex]\lim_{x \to \infty}=\frac{3}{ \sqrt{4}}[/tex]
[tex]\lim_{x \to \infty}=\frac{3}{2}[/tex]
Keep in mind that since the denominator is irrational, we also need to check the left horizontal asymptote:
[tex]f(x)=\frac{3x-2}{\sqrt{4x^2+5}}[/tex]
[tex]\lim_{x \to -\infty}=\frac{3x-2}{\sqrt{4x^2+5}}[/tex]
[tex]\lim_{x \to -\infty}=\frac{3-\frac{2}{x}}{\frac{1}{x} \sqrt{4x^2+5}}[/tex]
[tex]\lim_{x \to -\infty}=\frac{3-\frac{2}{x}}{-\sqrt{\frac{1}{x^2}}\sqrt{4x^2+5}}}[/tex]
[tex]\lim_{x \to -\infty}=\frac{3-\frac{2}{x}}{-\sqrt{4+\frac{5}{x^2}}}[/tex]
[tex]\lim_{x \to -\infty}=\frac{3}{-\sqrt{4}}[/tex]
[tex]\lim_{x \to -\infty}=\frac{3}{-2}[/tex]
[tex]\lim_{x \to -\infty}=-\frac{3}{2}[/tex]
Therefore, the horizontal asymptotes of the function are [tex]y=-\frac{3}{2}[/tex] and [tex]y=\frac{3}{2}[/tex].
