Biology Forums - Study Force

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 ) = 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. Show
work 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 explanations
a. 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 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))

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

False

This is not true for function f. It is valid only If f is continuous and monotonically increasing in (a, b).

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: \(D=\left\{x\left|x>\sqrt{12},x<-\sqrt{12}\right|\right\}\)

\(f\left(x\right)=4x^3+x+3\)

\(f\left(-1\right)=4\left(-1\right)^3+\left(-1\right)+3=-2\)

\(f\left(0\right)=4\left(0\right)^3+\left(0\right)+3=+3\)

\(f\left(1\right)=4\left(1\right)^3+\left(1\right)+3=8\)

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.

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.

maybe >= 2√3?

Do they expect use to actually graph these? Naw.

Here's a brighter way, use L'Hopital's rule.

a) Limit is 2
b) Limit is 0
c) Limit is 0
d) Limit is infinity

https://www.youtube.com/watch?v=Pi2OfBG_1s8

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.

"analytically" makes no sense, I think they meant algebraically.

\(\frac{2x}{12x+6}=\frac{\frac{2x}{2x}}{\frac{12x}{2x}+\frac{6}{2x}}=\lim \frac{1}{6+\frac{3}{x}}=\frac{1}{6}\ \)

\(\lim \left(8+\frac{5}{x^2}\right)=8+\lim \left(\frac{5}{x^2}\right)=8+0=8\)

\(\frac{x^3+2}{x^3+\sqrt{4x^6+3}}=\frac{\frac{x^3}{x^3}+\frac{2}{x^3}}{\frac{x^3}{x^3}+\frac{\sqrt{x^6\left(4+\frac{3}{x^6}\right)}}{x^3}}=\lim \left(\frac{1+\frac{2}{x^3}}{1+\frac{x^3\sqrt{4+\frac{3}{x^6}}}{x^3}}\right)=\frac{1}{1+\sqrt{4+\frac{3}{x^6}}}=\frac{1}{1+\sqrt{4}}=\frac{1}{3}\)

This required some mad algebra skills.

Horizontal 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)

https://www.youtube.com/watch?v=whtaqcS_Ei4

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.