Functions - Pre-Calculus

Card 0 of 644

Question

Which of the following is not true about a field. (Note: the real numbers $\small \mathbb{R}$ is a field)

Answer

It is not the case that for any element $\small a$ in a field, there is another one $\small b$ such that their product is $\small 1$ . Take $\small 0$ in the real numbers. Multiply $\small 0$ by any number and you get $\small 0$ , so you will never get $\small 1$ . This is true for any field that has more than 1 element.

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Question

What is the inverse function of

?

Answer

To find the inverse function of

we replace the with and vice versa.

So

Now solve for

$\frac{3y}{3}=\frac{x-5}{3}$

$y=\frac{x-5}{3}$

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Question

What is the inverse of $\left(\frac{1}{2},3\right)$ ?

Answer

When trying to find the inverse of a point, switch the x and y values.

So, $\left(\frac{1}{2},3\right)\rightarrow \left(3, \frac{1}{2} \right )$

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Question

Find the inverse of the following function:

Answer

In order to find the inverse of the function, we need to switch the x- and y-variables.

After switching the variables, we have the following:

Now solve for the y-variable. Start by subtracting 10 from both sides of the equation.

Divide both sides of the equation by 4.

$\frac{x-10}{4}=\frac{4y}{4}$

Rearrange and solve.

$y=\frac{x-10}{4}$

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Question

Find the inverse of the function.

$f(x)= \sin{\frac{x^2}{10}}$

Answer

To find the inverse function, first replace with :

$y= \sin{ \frac{x^2}{10} }$

Now replace each with an and each with a :

$x= \sin{ \frac{y^2}{10} }$

Solve the above equation for :

$\arcsin{x}= \frac{y^2}{10}$

$\sqrt{10 \arcsin{x}} = y$

Replace with $f^{-1}(x)$ . This is the inverse function:

$f^{-1}(x)=\sqrt{ 10 \arcsin{x} }$

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Question

Find the inverse of the function.

$g(x)= \frac{x+3}{2x-4}$

Answer

To find the inverse function, first replace with :

$y= \frac{x+3}{2x-4}$

Now replace each with an and each with a :

$x= \frac{y+3}{2y-4}$

Solve the above equation for :

$y= \frac{4x+3}{2x-1}$

Replace with $g^{-1}(x)$ . This is the inverse function:

$g^{-1}(x)= \frac{4x+3}{2x-1}$

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Question

Find the inverse of the function .

Answer

To find the inverse of , interchange the and terms and solve for .

$\frac{x-1}{2}=y^2$

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

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Question

What point is the inverse of the ?

Answer

When trying to find the inverse of a point, switch the x and y values.

So $\rightarrow (3,2)$

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Question

Let

$y=\left{\begin{matrix} 2x&x\leq 0\ x+9& x>0 \end{matrix}\right.$

What does equal when ?

Answer

Because 3>0 we plug the x value into the bottom equation.

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Question

Let

$y=\left{\begin{matrix} x-9&x\leq 0\ x+9& x>0 \end{matrix}\right.$

What does equal when ?

Answer

Because $0\leq0$ we use the first equation.

Therefore, plugging in x=0 into the above equation we get the following,

.

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Question

Determine the value of if the function is

$f(x)=\left{\begin{matrix} x& x<3\ \pi&x=3 \ 4&x>3 \end{matrix}\right.$

Answer

In order to determine the value of of the function we set

The value comes from the function in the first row of the piecewise function, and as such

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Question

Determine the value of if the function is

$f(x)=\left{\begin{matrix} \sqrt{x}& 0\leq x<4\ \3&x=4 \ e^{-x}&x>4 \end{matrix}\right.$

Answer

In order to determine the value of of the function we set

The value comes from the function in the first row of the piecewise function, and as such

$f(1)=\sqrt{1}=1$

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Question

For the function defined below, what is the value of when ?

$f(x)=\left{\begin{matrix} -x+1&\text{ for }x<1\ x^{2}-2&\text{ for }1< x< 3\ 2^{x}&\text{ for }x\geq 3 \end{matrix}$

Answer

Evaluate the function for . Based on the domains of the three given expressions, you would use $2^{x}$ , since is greater than or equal to .

$2^{x}=2^{3}=8$

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Question

What is the domain of the following function:

$f(x)=\frac{1}{1+\sqrt x}$

Answer

Note that in the denominator, we need to have $x\ge 0$ to make the square root of x defined. In this case $1+\sqrt x$ is never zero. Hence we have no issue when dividing by this number. Therefore the domain is the set of real numbers that are $\ge 0$

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Question

Find the range of the following function:

Answer

Every element of the domain has as image 7.This means that the function is constant . Therefore,

the range of f is :{7}.

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Question

Find the domain of the following function f(x) given below:

$f(x)=\sqrt{ x^3-1}$

Answer

. Since $(x^2-x+1)\ >0$ for all real numbers. To make the square root positive we need to have $x-1\ge 0$ .

Therefore the domain is :

${x\in\mathbb{R}: x\ge 1}=[1,\infty)$

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Question

What is the range of :

Answer

We know that $-1\le cos(x)\le 1$ . So $-2\le 2cos(x)\le 2$ .

Therefore:

$1-2\le 1+2cos(x)\le 1+2$ .

This gives:

$-1\le 1+2cos(x)\le 3$ .

Therefore the range is:

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Question

Find the range of f(x) given below:

Answer

Note that: we can write f(x) as :

.

Since,

$(x+1)^2\ge 0 $ for all reals.$

Therefore,

$(x+1)^2+2\ge 2.$

So the range is $[2,\infty)$

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Question

What is the range of :

Answer

We have $-1\le cos(3x)\le 1$ .

Adding 7 to both sides we have:

$6\le cos(3x)\le 8$ .

Therefore $6\le f(x)\le 8$ .

This means that the range of f is

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Question

Find the domain of the following function:

$f(x)=\sqrt{121-x}$

Answer

The part inside the square root must be positive. This means that we must be $\ge 0$ . Thus

$121-x\ge0$ . Adding -121 to both sides gives $-x\ge -121$ . Finally multiplying both sides by (-1) give:

$x\le 121$ with x reals. This gives the answer.

Note: When we divide by a negative we need to flip our sign.

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