SAT Math - Algebraic Functions

If f(x) = _x_2 – 5 for all values x and f(a) = 4, what is one possible value of a?

Explanation

Using the defined function, f(a) will produce the same result when substituted for x:

f(a) = _a_2 – 5

Setting this equal to 4, you can solve for a:

_a_2 – 5 = 4

_a_2 = 9

a = –3 or 3

What is the range of the function y = _x_2 + 2?

all real numbers

y ≥ 2

{–2, 2}

{2}

undefined

Explanation

The range of a function is the set of y-values that a function can take. First let's find the domain. The domain is the set of x-values that the function can take. Here the domain is all real numbers because no x-value will make this function undefined. (Dividing by 0 is an example of an operation that would make the function undefined.)

So if any value of x can be plugged into y = _x_2 + 2, can y take any value also? Not quite! The smallest value that y can ever be is 2. No matter what value of x is plugged in, y = _x_2 + 2 will never produce a number less than 2. Therefore the range is y ≥ 2.

If f(x) = _x_2 – 5 for all values x and f(a) = 4, what is one possible value of a?

Explanation

Using the defined function, f(a) will produce the same result when substituted for x:

f(a) = _a_2 – 5

Setting this equal to 4, you can solve for a:

_a_2 – 5 = 4

_a_2 = 9

a = –3 or 3

What is the range of the function y = _x_2 + 2?

all real numbers

y ≥ 2

{–2, 2}

{2}

undefined

Explanation

An outpost has the supplies to last 2 people for 14 days. How many days will the supplies last for 7 people?

$\dpi{100} \small 4$

$\dpi{100} \small 7$

$\dpi{100} \small 5$

$\dpi{100} \small 10$

$\dpi{100} \small 9$

Explanation

Supplies are used at the rate of $\dpi{100} \small \frac{Supplies}{Days\times People}$ .

Since the total amount of supplies is the same in either case, $\dpi{100} \small \frac{1}{14\times 2}=\frac{1}{7\times \ (&hash;\ of\ days)}$ .

Solve for days to find that the supplies will last for 4 days.

An outpost has the supplies to last 2 people for 14 days. How many days will the supplies last for 7 people?

$\dpi{100} \small 4$

$\dpi{100} \small 7$

$\dpi{100} \small 5$

$\dpi{100} \small 10$

$\dpi{100} \small 9$

Explanation

Supplies are used at the rate of $\dpi{100} \small \frac{Supplies}{Days\times People}$ .

Since the total amount of supplies is the same in either case, $\dpi{100} \small \frac{1}{14\times 2}=\frac{1}{7\times \ (&hash;\ of\ days)}$ .

Solve for days to find that the supplies will last for 4 days.

Explanation

When we multiply a function by a constant, we multiply each value in the function by that constant. Thus, 2f(x) = 4x + 12. We then subtract g(x) from that function, making sure to distribute the negative sign throughout the function. Subtracting g(x) from 4x + 12 gives us 4x + 12 - (3x - 3) = 4x + 12 - 3x + 3 = x + 15. We then add 2 to x + 15, giving us our answer of x + 17.

Explanation

The cost of a cell phone plan is $40 for the first 100 minutes of calls, and then 5 cents for each minute after. If the variable x is equal to the number of minutes used for calls in a month on that cell phone plan, what is the equation f(x) for the cost, in dollars, of the cell phone plan for calls during that month?

f(x) = 40 + 5x

f(x) = 40 + 0.5x

f(x) = 40 + 0.05x

f(x) = 40 + 0.5(x - 100)

f(x) = 40 + 0.05(x - 100)

Explanation

40 dollars is the constant cost of the cell phone plan, regardless of minute usage for calls. We then add 5 cents, or 0.05 dollars, for every minute of calls over 100. Thus, we do not multiply 0.05 by x, but rather by (x - 100), since the 5 cent charge only applies to minutes used that are over the 100-minute barrier. For example, if you used 101 minutes for calls during the month, you would only pay the 5 cents for that 101st minute, making your cost for calls $40.05. Thus, the answer is 40 + 0.05(x - 100).

Let $f(x)$ and $g(x)$ be functions such that $f(x)=\frac{1}{x-3}$ , and $f(g(x))=g(f(x))=x$ . Which of the following is equal to $g(x)$ ?

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

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

$\frac{x}{x-3}$

$\frac{4}{x}$

$\frac{x}{x+3}$

Explanation

If $\dpi{100} h(x)$ and $\dpi{100} k(x)$ are defined as inverse functions, then $\dpi{100} h(k(x))=k(h(x))=x$ . Thus, according to the definition of inverse functions, $\dpi{100} f(x)$ and $\dpi{100} g(x)$ given in the problem must be inverse functions.

If we want to find the inverse of a function, the most straighforward method is usually replacing $\dpi{100} f(x)$ with $\dpi{100} y$ , swapping $\dpi{100} y$ and $\dpi{100} x$ , and then solving for $\dpi{100} y$ .

We want to find the inverse of $f(x)=\frac{1}{x-3}$ . First, we will replace $\dpi{100} f(x)$ with $\dpi{100} y$ .

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

Next, we will swap $\dpi{100} x$ and $\dpi{100} y$ .

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

Lastly, we will solve for $\dpi{100} y$ . The equation that we obtain in terms of $\dpi{100} x$ will be in the inverse of $\dpi{100} f(x)$ , which equals $\dpi{100} g(x)$ .

We can treat $x = \frac{1}{y-3}$ as a proportion, $\frac{x}{1} = \frac{1}{y-3}$ . This allows us to cross multiply and set the results equal to one another.