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5 months ago
 Determine whether or not of the following statements are true or false, and give explanations or cou 1. Determine whether or not of the following statements are true or false, andgive explanations or counterexamples for each.If a < b and f ( a ) < L < f ( b ), then there is some value of c in ( a , b ) for which f ( c ) = L 2.Determine the interval(s) on which the function f(x) = √(4x^2-48) is continuous. Show work and give exact answers. Be sure to consider le- and right-continuity at the endpoints.3.  Use the Intermediate Value Theorem to show that the equation 4x^3 + x + 3 = 0 has a  solution on the interval (− 1, 1) . Show all work and explanations.4. Determine all the points at which the function f(x) = csc x has discontinuities. Showwork and explanations.5.Evaluate the following graphically , showing accurate sketches of graphs as your work.a. lim x→∞2x/(x+1)b. lim x→∞2x/(x^2 +1)c. lim x→∞2^(−x)d. lim x→∞ (2ln x)6.Evaluate the following analytically (i.e., by hand), showing all work or explanationsa. lim x→∞ 2x/(12x+6)b. lim x→∞ (8 +5/x^2 )7. Find the following analytically (i.e., by hand), showing all work or explanationsa. lim x→∞(x^3 +2)/(x^3+sqrt(4x^6+3))b. lim x→−∞(x^3 +2)/(x^3+sqrt(4x^6+3))9. Determine the horizontal asymptote(s) of g(x) = (3x^3 +16x^2 +16x)/ ∣x∣ analytically (i.e., by hand) showing all work or explanations Read 120 times 10 Replies
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5 months ago
 Quote from: chansamhua (5 months ago)1. Determine whether or not of the following statements are true or false, and give explanations or counterexamples for each. If a < b and f ( a ) < L < f ( b ), then there is some value of c in ( a , b ) for which f ( c ) = LFalseThis is not true for function f. It is valid only If f is continuous and monotonically increasing in (a, b).
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5 months ago
 Quote from: chansamhua (5 months ago)2.Determine the interval(s) on which the function f(x) = √(4x^2-48) is continuous. Show work and give exact answers. Be sure to consider le- and right-continuity at the endpoints.For this, find the domain.$$\sqrt{4x^2-48}$$$$4x^2-48\ >0$$$$x^2>\frac{48}{4}$$$$x^2>12$$$$x>\pm \sqrt{12}$$The domain is: $$D=\left\{x\left|x>\sqrt{12},x<-\sqrt{12}\right|\right\}$$
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5 months ago
 Quote from: chansamhua (5 months ago)3.  Use the Intermediate Value Theorem to show that the equation 4x^3 + x + 3 = 0 has a  solution on the interval (− 1, 1) . Show all work and explanations.$$f\left(x\right)=4x^3+x+3$$$$f\left(-1\right)=4\left(-1\right)^3+\left(-1\right)+3=-2$$$$f\left(0\right)=4\left(0\right)^3+\left(0\right)+3=+3$$$$f\left(1\right)=4\left(1\right)^3+\left(1\right)+3=8$$It goes from negative to positive to negative when evaluated at -1 and 3 so the Intermediate Value Theorem guarantees that there exists a number c1 between these intervals so that f(c1) = 0. Remember, all polynomials are continuous. So we conclude that the equation has at least one solution on the intervals.
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5 months ago
 Quote from: chansamhua (5 months ago)4. Determine all the points at which the function f(x) = csc x has discontinuities. Show work and explanations.This function is the reciprocal function for $$\frac{1}{\sin \left(x\right)}$$. Fractions cannot have a denominator of 0.$$\sin \left(x\right)\ne 0$$When does sin(x) equal 0.When $$x=0°$$, when $$x=180°$$. That's where it's discontinuous.
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5 months ago
 Quote from: bio_man (5 months ago)Determine the interval(s) on which the function f(x) = √(4x^2-48) is continuous. Show work and give exact answers. Be sure to consider le- and right-continuity at the endpoints.maybe >= 2√3?
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5 months ago
 Quote from: chansamhua (5 months ago)5.Evaluate the following graphically , showing accurate sketches of graphs as your work. a. lim x→∞2x/(x+1) b. lim x→∞2x/(x^2 +1) c. lim x→∞2^(−x) d. lim x→∞ (2ln x)Do they expect use to actually graph these? Naw.Here's a brighter way, use L'Hopital's rule.a) Limit is 2b) Limit is 0c) Limit is 0d) Limit is infinity
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5 months ago
 Quote from: chansamhua (5 months ago) Quote from: bio_man (5 months ago)Determine the interval(s) on which the function f(x) = √(4x^2-48) is continuous. Show work and give exact answers. Be sure to consider le- and right-continuity at the endpoints. maybe >= 2√3? No, $$\sqrt{12}=\sqrt{4\cdot 3}=\pm 2\sqrt{3}$$So if you want, write:$$x>2\sqrt{3}$$ and $$x<-2\sqrt{3}$$. Those are your only two options, as proven.
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5 months ago
 Quote from: chansamhua (5 months ago)6.Evaluate the following analytically (i.e., by hand), showing all work or explanations a. lim x→∞ 2x/(12x+6) b. lim x→∞ (8 +5/x^2 )"analytically" makes no sense, I think they meant algebraically.$$\frac{2x}{12x+6}=\frac{\frac{2x}{2x}}{\frac{12x}{2x}+\frac{6}{2x}}=\lim \frac{1}{6+\frac{3}{x}}=\frac{1}{6}\$$$$\lim \left(8+\frac{5}{x^2}\right)=8+\lim \left(\frac{5}{x^2}\right)=8+0=8$$
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5 months ago
 Quote from: chansamhua (5 months ago)7. Find the following analytically (i.e., by hand), showing all work or explanations a. lim x→∞(x^3 +2)/(x^3+sqrt(4x^6+3)) b. lim x→−∞(x^3 +2)/(x^3+sqrt(4x^6+3))$$\frac{x^3+2}{x^3+\sqrt{4x^6+3}}=\frac{\frac{x^3}{x^3}+\frac{2}{x^3}}{\frac{x^3}{x^3}+\frac{\sqrt{x^6\left(4+\frac{3}{x^6}\right)}}{x^3}}=\lim \left(\frac{1+\frac{2}{x^3}}{1+\frac{x^3\sqrt{4+\frac{3}{x^6}}}{x^3}}\right)=\frac{1}{1+\sqrt{4+\frac{3}{x^6}}}=\frac{1}{1+\sqrt{4}}=\frac{1}{3}$$This required some mad algebra skills.
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5 months ago
 Quote from: chansamhua (5 months ago)9. Determine the horizontal asymptote(s) of g(x) = (3x^3 +16x^2 +16x)/ ∣x∣ analytically (i.e., by hand) showing all work or explanationsHorizontal asymptotes technically means they want the range, what's the highest or lowest value of value that exists, hence the domain. Luckily, at the numerator, it's a polynomial, so no issues there. At the bottom, we have |x|, you could distribute that to each numerator term like this to form a polynomial:$$g\left(x\right)=\frac{3x^3+16x^2+16x}{\left|x\right|}=\frac{3x^3}{\left|x\right|}+\frac{16x^2}{\left|x\right|}+\frac{16x}{\left|x\right|}=-3x^2-16x-16$$To find the maximum/minimum, use $$h=-\frac{b}{2a}$$, where h = x coordinate of the vertex. (shown below)You should get $$\left(-2\ \frac{2}{3},5\ \frac{1}{3}\right)$$. Therefore, the maximum value or horizontal asymptote is $$5\ \frac{1}{3}$$. Interestingly, without the absolute at the bottom, it would have been a minimum at $$-5\ \frac{1}{3}$$, and the parabola would be facing up.
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