Algebra - How to find f(x)

Explanation

$2\times ×36+6+2$

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:

$\small f(x)=2x+17$ .

Find $\small f(-3)$ .

$\small 11$

$\small 23$

$\small 34$

$\small 16$

Explanation

Finding $\small f(-3)$ means plugging in $\small -3$ in place of x into the function $\small f(x)$ . In this case, we are told that $\small f(x)=2x+17$ , so we are multiplying $\small -3$ by 2 and then adding 17:

$\small \small f(-3)=2(-3)+17$ order of opperations tells us to first multiply:

$\small f(-3)=-6+17$ and now add:

$\small f(-3)= 11$

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 3.

Each of the following 4 sets defines a relationship between and . Which of these four sets defines a one-to-one function:

A = $\left { \left ( 1,2 \right )\left ( 2,3 \right )\left ( 3,4 \right )\left ( 4,5 \right ) \right }$

B= $\left { \left ( 1,3 \right )\left ( 2,3 \right )\left ( 3,5 \right )\left ( 4,5 \right ) \right }$

C = $\left { \left ( 1,2 \right )\left ( 2,3 \right )\left ( 2,4 \right )\left ( 2,5 \right )\left ( 6,7\right )\left\right }$

D = $\left { \left ( 1,7 \right )\left ( 2,7 \right )\left ( 3,7 \right )\left ( 4,7 \right ) \right }$

Set A

Set B

Set C

Set D

Set A and Set B

Explanation

Only in set A one can see that there is an unique value of for each value of and similarly each of the values maps into one and only one value. Hence set A must define a one-to-one function.