Algebraic Functions

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Algebra › Algebraic Functions

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Explanation

$2\times ×36+6+2$

If an object is hung on a spring, the elongation of the spring varies directly as the mass of the object. A 20 kg object increases the length of a spring by exactly 7.2 cm. To the nearest tenth of a centimeter, by how much does a 32 kg object increase the length of the same spring?

$11.5 \textrm{ cm}$

$4.5 \textrm{ cm}$

$18.4 \textrm{ cm}$

$9.1\textrm{ cm}$

$8.4 \textrm{ cm}$

Explanation

Let be the mass of the weight and the elongation of the spring. Then for some constant of variation ,

We can find by setting from the first situation:

$7.2 = k \cdot 20$

$k = 7.2 \div 20 = 0.36$

In the second situation, we set and solve for :

$L = 0.36 \cdot32 =11.52$

which rounds to 11.5 centimeters.

Given a function $f(x)=x^{3}-9x+3$ , what is ?

Explanation

Given a function $f(x)=x^{3}-9x+3$ , we can plug in to get

$\ f(3)=(3)^{3}-9(3)+3 \=27-27+3\=3$ .

If , then solve

Explanation

We know . When solving a function, we substitute the value of x into the function. In other words, anywhere we see an x, we will replace it with -4.

If , solve for if $f(\frac{1}{7})$ .

$-\frac{6}{7}$

$-7\frac{1}{6}$

$7\frac{1}{6}$

Explanation

We are given an equation that is a function of x. Substitute the given fraction and replace it with the variable.

$f(\frac{1}{7}) = 7(\frac{1}{7})-6$

Simplify the right side.

$7(\frac{1}{7})-6 = 1-6 = -5$

The answer is:

Find the inverse of the following algebraic function:

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

$y=\frac{1}{x}+3$

$y=\frac{1}{x}-3$

$y=x+\frac{1}{3}$

$y=x-\frac{1}{3}$

Explanation

To find the inverse, switch the placement of the and variables:

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

$y=\frac{1}{x-3}$

$x=\frac{1}{y-3}$

Next, should be isolated, providing the inverse function:

$\frac{y-3}{1}\cdot x=\frac{1}{y-3}\cdot \frac{y-3}{1}$

$\frac{xy}{x}=\frac{1+3x}{x}$

$y=\frac{1+3x}{x}$

$y=\frac{1}{x}+3$

varies inversely as the square root of . If , then . Find if (nearest tenth, if applicable).

Explanation

The variation equation is $A = \frac{K}{\sqrt{x}}$ for some constant of variation .

Substitute the numbers from the first scenario to find :

$45= \frac{K}{\sqrt{15}}$

$K = 45 \sqrt{15}$

The equation is now $A = \frac{45 \sqrt{15}}{\sqrt{x}}$ .

If , then

$A = \frac{45 \sqrt{15}}{\sqrt{30}} \approx 31.8$

Find the inverse:

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

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

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

Explanation

To find the inverse, first interchange the x and y variables in the equation.

Solve for y. Add 6 on both sides.

Simplify.

Divide by four on both sides.

$\frac{x+6}{4}=\frac{4y}{4}$

Simplify both sides.

$y=\frac{x+6}{4} = \frac{1}{4}x + \frac{6}{4}= \frac{1}{4}x + \frac{3}{2}$

The inverse is: $y= \frac{1}{4}x + \frac{3}{2}$

Explanation

$2\times ×36+6+2$

is a one-to-one function specified in terms of a set of $\left ( x,y \right )$ coordinates:

A = $\left { \left ( 0,0 \right ), \left ( 1,2\right ),\left ( 2,4 \right ), \left ( 3,6 \right )\right }$

Which one of the following represents the inverse of the function specified by set A?

B = $\left { \left ( 0,0 \right ), \left ( 0,1 \right ), \left ( 0,2 \right ), \left ( 0,3 \right ) \right }$

C = $\left { \left ( 0,0 \right ), \left ( 2,1 \right ), \left ( 4,2 \right ), \left ( 6,3 \right ) \right }$

D = $\left { \left ( 1,1 \right ), \left ( 2,1 \right ), \left ( -2,1 \right ), \left ( 3,1 \right ) \right }$

E = $\left { \left ( -2,1 \right ), \left ( -2,5 \right ), \left ( -2,3 \right ), \left ( -2, -2 \right ) \right }$

F = $\left { \left ( 2,0 \right ), \left ( 2,-3 \right ),\left ( 2,5 \right ), \left ( 2,-7 \right ) \right }$

Set C

Set B

Set D

Set E

Set F

Explanation

The set A is an one-to-one function of the form

One can find $f^{-1}\left ( x \right )$ by interchanging the and coordinates in set A resulting in set C.

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