Math - Pre-Calculus | Practice Hub

Simplify the following polynomial function:

$ab^{-2}(a^{2}b + a^{-1}b^{3}-a^{3}b^{-1})$

$\frac{a^{3}}{b}+b-\frac{a^{4}}{b^{3}}$

$\frac{a^{3}}{b}-\frac{a^{4}}{b^{3}}$

$\frac{a^{3}}{b}+\frac{a^{4}}{b^{3}}$

$\frac{a^{3}}{b}+b^{2}-\frac{a^{4}}{b^{3}}$

$\frac{a^{3}}{b}-b+\frac{a^{4}}{b^{3}}$

Explanation

First, multiply the outside term with each term within the parentheses:

$ab^{-2}(a^{2}b + a^{-1}b^{3}-a^{3}b^{-1})$

$a^{3}b^{-1} + b-a^{4}b^{-3}$

Rearranging the polynomial into fractional form, we get:

$\frac{a^{3}}{b}+b-\frac{a^{4}}{b^{3}}$

What is the domain of the function below:

$f(x)=\frac{1}{\sqrt[4]{x^{2}+1}-\frac{1}{2}}$

** $x>\frac{1}{2}$

$\mathbb{R}$

Explanation

The domain is defined as the set of possible values for the x variable. In order to find the impossible values of x, we should:

a) Set the equation under the radical equal to zero and look for probable x values that make the expression inside the radical negative:

There is no real value for x that will fit this equation, because any real value square is a positive number i.e. cannot be a negative number.

b) Set the denominator of the fractional function equal to zero and look for probable x values:

$\sqrt[4]{x^2+1}-\frac{1}{2}=0$

Now we can solve the equation for x:

$x^2+1=(\frac{1}{2})^4$

There is no real value for x that will fit this equation.

The radical is always positive and denominator is never equal to zero, so the f(x) is defined for all real values of x. That means the set of all real numbers is the domain of the f(x) and the correct answer is $\mathbb{R}$ .

Alternative solution for the second part of the solution:

After figuring out that the expression under the radical is always positive (part a), we can solve the radical and therefore denominator for the least possible value (minimum value). Setting the x value equal to zero will give the minimum possible value for the denominator.

$\sqrt[4]{x^2+1}-\frac{1}{2}=$

$\sqrt[4]{0+1}-\frac{1}{2}=$

$\frac{1}{2}>0$

That means the denominator will always be a positive value greater than 1/2; thus it cannot be equal to zero by setting any real value for x. Therefore the set of all real numbers is the domain of the f(x).

What is the domain of the function below?

$f(x)=\frac{1-2x}{\sqrt{x^2-x+\frac{1}{4}}}$

$x eq \frac{1}{2}$

$x eq -\frac{1}{2}$

$\mathbb{N}$

Explanation

The domain is defined as the set of all values of x for which the function is defined i.e. has a real result. The square root of a negative number isn't defined, so we should find the intervals where that occurs:

$x^2-x+\frac{1}{4} <0 \Rightarrow (x-\frac{1}{2})^2<0$

The square of any number is positive, so we can't eliminate any x-values yet.

If the denominator is zero, the expression will also be undefined.

Find the x-values which would make the denominator 0:

$(x-\frac{1}{2})^2=0\Rightarrow (x-\frac{1}{2})=0\Rightarrow x=\frac{1}{2}$

Therefore, the domain is $x eq \frac{1}{2}$ .

$\textup{Find: }\lim_{x \to \infty }\frac{2x^{2}-7x}{x^{2}+3}$

$\infty$

Explanation

$\textup{As } x \to\infty,\textup{all but the terms with the highest exponents become insignificant.}$

$\textup{As }x^{2} \textup{ is the highest on top and bottom, evaluate the coefficients:}} \frac{2}{1}=2$

Calculate $\lim_{x\rightarrow \infty }\left ( x^{2} \sin \frac{1}{ x^{2}-1} - \sin \frac{1}{ x^{2}-1} \right )$ .

$\sin 1$

$\infty$

The limit does not exist.

Explanation

This can be rewritten as follows:

$\lim_{x\rightarrow \infty }\left ( x^{2} \sin \frac{1}{ x^{2}-1} - \sin \frac{1}{ x^{2}-1} \right )$

$= \lim_{x\rightarrow \infty }\left ( x^{2}\cdot \sin \frac{1}{ x^{2}-1} -1\cdot \sin \frac{1}{ x^{2}-1} \right )$

$= \lim_{x\rightarrow \infty }\left [\left ( x^{2}-1 \right ) \cdot \sin \frac{1}{ x^{2}-1} \right]$

$= \lim_{x\rightarrow \infty } \frac{ \sin \frac{1}{ x^{2}-1} }{\frac{1}{ x^{2}-1}}$

We can substitute $u = \frac{1}{ x^{2}-1}$ , noting that as $x\rightarrow \infty$ , $u \rightarrow 0$ :

$= \lim_{u\rightarrow 0} \frac{ \sin u }{u} = 1$ , which is the correct choice.

What are the solutions to $y = x^{2} + x -6$ ?

Explanation

When we are looking for the solutions of a quadratic, or the zeroes, we are looking for the values of such that the output will be zero. Thus, we first factor the equation.