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SAT Math : How to find f(x)

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Example Question #102 :Algebraic Functions

f(x)=2x+6

g(x)=3x-3

2f(x)-g(x)+2=?

Possible Answers:

x + 11

x + 5

x + 17

5x + 11

7x + 17

Correct answer:

x + 17

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 ofx + 17.

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Example Question #103 :Algebraic Functions

Given the following functions, evaluate b(4) - w(5) .

b(x)=x^2+2x

w(x)=x^2-2x

Possible Answers:

Correct answer:

Explanation:

b(4)=4^2+2(4)=24

w(5)=5^2-2(5)=15

b(4)-w(5)=24-15=9

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Example Question #104 :Algebraic Functions

The point (3,w) is one of the points of intersection of the graphs of f(x) and g(x) . Given that f(x)=x^2-2 and g(x)=-x^2+u , find the value of.

Possible Answers:

Correct answer:

Explanation:

Since it’s given than f(3) = w , we know that w=3^2-2=7 . We now know that (3, 7) is a point of intersection. Therefore, g(3) = 7 as well. In other words, 7=-3^2+u , so u=16 .

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Example Question #105 :Algebraic Functions

Consider these functions:

b(x)=x^2+2x

w(x)=x^2-2x

Which of the following is equivalent to w(e-1) ?

Possible Answers:

b(e)-2e+1

b(e)-4e+3

b(e)+6e-3

b(e)+4e

b(e)-6e+3

Correct answer:

b(e)-6e+3

Explanation:

w(e-1)=(e-1)^2-2(e-1)

=e^2-2e+1-2e+2

=e^2-4e+3

=e^2+2e-6e+3

=b(e)-6e+3

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Example Question #106 :Algebraic Functions

If $f(t)=6+(4-t)^2$ , what is the smallest possible value of $f(t)$ ?

Possible Answers:

Correct answer:

Explanation:

This equation describes a parabola whose vertex is located at the point (4, 6). No matter how large or small the value of t gets, the smallest that f(t) can ever be is 6 because the parabola is concave up. To prove this to yourself you can plug in different values of t and see if you ever get anything smaller than 6.

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Example Question #12 :Algebraic Functions

If $f(x)=x^{2}+3$ , then $f(x+h)=$ ?

Possible Answers:

$x^{2}+3+h$

$x^{2}+h^{2}$

$x^{2}+2xh+h^{2}+3$

$x^{2}+2xh+h^{2}$

$x^{2}+h^{2}+3$

Correct answer:

$x^{2}+2xh+h^{2}+3$

Explanation:

To find $f(x+h)$ when $f(x)=x^{2}+3$ , we substitute $(x+h)$ for $x$ in $f(x)$ .

Thus, $f(x+h)=(x+h)^{2}+3$ .

We expand $(x+h)^{2}$ to $x^{2}+xh+xh+h^{2}$ .

We can combine like terms to get $x^{2}+2xh+h^{2}$ .

We add 3 to this result to get our final answer.

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Example Question #101 :Algebraic Functions

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 $g(x)$ ?

Possible Answers:

$\frac{4}{x}$

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

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

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

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

Correct answer:

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

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.

$x(y-3)= 1$

We want to get y by itself, so let's divide both sides by x.

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

Next, we will add 3 to both sides.

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

To combine the right side, we will need to rewrite 3 so that it has a denominator of $\dpi{100} x$ .

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

The answer is $\frac{3x+1}{x}$ .

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Example Question #108 :Algebraic Functions

$\small \textup{The function}\, f(x)\, \textup{is defined as }f\left ( x\right )=5x-2$ .

$\small \small f\left ( 4\right )-f\left ( 2\right )=\,?$

Possible Answers:

$\small 8$

$\small 10$

$\small 2$

$\small 14$

$\small 6$

Correct answer:

$\small 10$

Explanation:

$\small f\left ( x\right )=5x - 2$

$\small f\left ( 4\right )=5\left ( 4\right )-2 = 18$

$\small f\left ( 2\right )=5\left ( 2\right )-2 = 8$

$\small 18-8 = 10$

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Example Question #2811 :Sat Mathematics

Let the functionfbe defined byf(x)=x-t. Iff(12)=4, what is the value off(0.5*t)?

Possible Answers:

Correct answer:

Explanation:

First we substitute in 12 for x and set the equation up as 12-t=4. We then get t=8, and substitute that for t and get f(0.5*8), giving us f(4). Plugging 4 in for x, and using t=8 that we found before, gives us:

f(4) = 4 - 8 = -4

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Example Question #71 :Algebraic Functions

What is the value of the function f(x) = 6x²+ 16x – 6 when x = –3?

Possible Answers:

–12

–108

Correct answer:

Explanation:

有两种方法可以解决这个问题。第一个way just involves plugging in –3 for x and solving 6〖(–3)〗²+ 16(–3) – 6, which equals 54 – 48 – 6 = 0. The second way involves factoring the polynomial to (6x – 2)(x + 3) and then plugging in –3 for x. The second way quickly shows that the answer is 0 due to multiplying by (–3 + 3).

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SAT Math : How to find f(x)

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #102 :Algebraic Functions

Example Question #103 :Algebraic Functions

Example Question #104 :Algebraic Functions

Example Question #105 :Algebraic Functions

Example Question #106 :Algebraic Functions

Example Question #12 :Algebraic Functions

Example Question #101 :Algebraic Functions

Example Question #108 :Algebraic Functions

Example Question #2811 :Sat Mathematics

Example Question #71 :Algebraic Functions

All SAT Math Resources

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