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Nic P.
Mathematics tutor for 6 years with background in mathematics education
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Pre-Calculus
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Question:

Find the domain and range for $f(x) = \sqrt{4 - x^2}$.

Nic P.

First let's find the domain. We recall that the "domain" is the set of numbers that a function is allowed to take as input. For a quick example, $0$ is not in the domain of the function $\frac{1}{x}$ because we are not allowed to divide by $0$. In a similar sense, when we look at the proposed question we should note that we cannot take square roots of negative numbers. This means that the quantity $(4-x^2)$ needs to be greater than (or equal to) $0$ for any $x$ value. So we should set up the simple inequality: $4-x^2 \geq 0$. Simple algebra yields that $|x| \leq 2$, which means that the domain is $-2 \leq x \leq 2$. The range is the set of numbers that $f(x)$ can be given any input in its domain. Essentially the range is the set of $y$-values you'd see on a graph if we graphed $f(x)$. To start, let's try the endpoints of the domain and see what happens. $f(-2) = \sqrt{4-(-2)^2} = 0$, and $f(2) = \sqrt{4 - 2^2} = 0$. So it looks like the graph hits $0$ at the end points of its domain. Certainly the range can't only be $0$? If we manipulate the given equation a little bit we can see what is happening. If we square both sides and move the $x$ over we get $f(x)^2 + x^ = 4$ which we should recognize as the formula for a circle with center at $(0,0)$ and radius $2$. Since the original equation had a square root, then we're only looking at the top-half of the circle, so the range is all the number on the y-axis between $0$ and $2$, ie. $0 \leq f(x) \leq 2$.

Trigonometry
TutorMe
Question:

Suppose $0 \leq \Theta \leq 2 \pi$, and $\sin(\Theta) = \frac{\sqrt{2}{2}}$. If $\cot(\Theta) < 0$ then which of the following is a possible value of $\Theta$? \beign[enumerate] \item[a.] $\frac{\pi}{6}$ \item[b.] $\frac{5 \pi}{6}$ \item[c.] $\frac{15 \pi}{6}$ \item[d.] $\frac{23 \pi}{6}$ \end[enumerate]

Nic P.

The answer is $\frac{5 \pi}{6}$. We first note that $\cot(\Theta) = \frac{\cos(\Theta)}{\sin(\Theta)}$. We're told that $\sin(\Theta)$ is positive and that $\cot(\Theta)$ is negative. This means that $\cos(\Theta)$ must be negative. $\cos()$ is negative in the second and third quadrants, but $\sin()$ is positive (given) in the first and second quadrants. Therefore the angle $\Theta$ must be in the second quadrant, and the only possible choice that fits this is choice b., $\Theta = \frac{5 \pi}{6}$.

Calculus
TutorMe
Question:

Estimate the error if the Taylor polynomial $P_3(x) = x - \frac{x^3}{6}$ is used to estimate $\sin(1)$.

Nic P.

We are given an estimate to the function $\sin(x)$ and $x=1$, so call $f(x) = \sin(x)$. Since we are given a third-degree polynomial to estimate $\sin(1)$ then the error will be bound, via Taylor's theorem, by the following: $|P_3(x) = \sin(x)| = R_3(x) = \frac{\sin(c)}{4!}x^4$ for some $c \in $0, \frac{\pi}{2}$$, since $f^{(4)}(x) = \sin(x)$. We know that $|\sin(c)| \leq 1$, thus we have that $|R_3(x)| \leq \frac{x^4}{4!}$. This implies that the approximate error in estimating $\sin(1)$ is $\frac{1^4}{4!} = \frac{1}{24}$.

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