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Example Questions

Example Question #101 : Algebraic Functions

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

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Example Question #12 : Algebraic Functions

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

Possible Answers:

–108

–12

Correct answer:

Explanation:

There are two ways to do this problem. The first 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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Example Question #11 : Algebraic Functions

Given the functions f(x) = 2x + 4 and g(x) = 3x – 6, what is f(g(x)) when x = 6?

Possible Answers:

192

144

Correct answer:

Explanation:

We need to work from the inside to the outside, so g(6) = 3(6) – 6 = 12.

Then f(g(6)) = 2(12) + 4 = 28.

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Example Question #12 : How To Find F(X)

A function f(x) = –1 for all values of x. Another function g(x) = 3x for all values of x. What is g(f(x)) when x = 4?

Possible Answers:

–1

–12

–3

Correct answer:

–3

Explanation:

We work from the inside out, so we start with the function f(x). f(4) = –1. Then we plug that value into g(x), so g(f(x)) = 3 * (–1) = –3.

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Example Question #342 : Algebra

What is f(–3) if f(x) = x² + 5?

Possible Answers:

–4

–14

Correct answer:

Explanation:

f(–3) = (–3)² + 5 = 9 + 5 = 14

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Example Question #21 : How To Find F(X)

For all values of x, f(x) = 7x² – 3, and for all values of y, g(y) = 2y + 9. What is g(f(x))?

Possible Answers:

14y² + 3

7y² – 3

14x² – 3

14x² + 3

2x + 9

Correct answer:

14x² + 3

Explanation:

The inner function f(x) is like our y-value that we plug into g(y).

g(f(x)) = 2(7x² – 3) + 9 = 14x² – 6 + 9 = 14x² + 3.

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Example Question #22 : How To Find F(X)

Find f(6) $\displaystyle f(6)$

$f(x)=|x^{2}+4x-127|$ $\displaystyle f(x)=|x^{2}+4x-127|$

Possible Answers:

$\displaystyle 67$

$\displaystyle -67$

$\displaystyle -136$

$\displaystyle -36$

$\displaystyle 36$

Correct answer:

$\displaystyle 67$

Explanation:

Simply plug 6 into the equation and don't forget the absolute value at the end.

$\displaystyle =36+24-127=-67$

absolute value = 67

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Example Question #23 : How To Find F(X)

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

Possible Answers:

$\dpi{100} \small 4$ $\displaystyle \dpi{100} \small 4$

$\dpi{100} \small 9$ $\displaystyle \dpi{100} \small 9$

$\dpi{100} \small 10$ $\displaystyle \dpi{100} \small 10$

$\dpi{100} \small 7$ $\displaystyle \dpi{100} \small 7$

$\dpi{100} \small 5$ $\displaystyle \dpi{100} \small 5$

Correct answer:

$\dpi{100} \small 4$ $\displaystyle \dpi{100} \small 4$

Explanation:

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

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

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

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Example Question #24 : How To Find F(X)

Worker $\dpi{100} \small A$ $\displaystyle \dpi{100} \small A$ can make a trinket in 4 hours, Worker $\dpi{100} \small B$ $\displaystyle \dpi{100} \small B$ can make a trinket in 2 hours. When they work together, how long will it take them to make a trinket?

Possible Answers:

$\dpi{100} \small \ 1 \frac{1}{3}\ hours$ $\displaystyle \dpi{100} \small \ 1 \frac{1}{3}\ hours$

$\dpi{100} \small 6\ hours$ $\displaystyle \dpi{100} \small 6\ hours$

$\dpi{100} \small 3\ hours$ $\displaystyle \dpi{100} \small 3\ hours$

$\dpi{100} \small \ 1 \frac{1}{2}\ hours$ $\displaystyle \dpi{100} \small \ 1 \frac{1}{2}\ hours$

$\dpi{100} \small \frac{1}{2}\ hour$ $\displaystyle \dpi{100} \small \frac{1}{2}\ hour$

Correct answer:

$\dpi{100} \small \ 1 \frac{1}{3}\ hours$ $\displaystyle \dpi{100} \small \ 1 \frac{1}{3}\ hours$

Explanation:

The rates are what needs to be added. Rate $\dpi{100} \small A$ $\displaystyle \dpi{100} \small A$ is $\dpi{100} \small \frac{1}{4}$ $\displaystyle \dpi{100} \small \frac{1}{4}$, or one trinket every 4 hours. Rate $\dpi{100} \small B$ $\displaystyle \dpi{100} \small B$ is $\dpi{100} \small \frac{1}{2}$ $\displaystyle \dpi{100} \small \frac{1}{2}$, one per two hours.

$\dpi{100} \small \frac{1}{4}+ \frac{1}{2}=\frac{3}{4}$ $\displaystyle \dpi{100} \small \frac{1}{4}+ \frac{1}{2}=\frac{3}{4}$, their combined rate in trinkets per hour.

Now invert the equation to get back to hours per trinket, which is what the question asks for: $\dpi{100} \small \frac{4}{3}\ or \ 1 \frac{1}{3}$ $\displaystyle \dpi{100} \small \frac{4}{3}\ or \ 1 \frac{1}{3}$

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Example Question #41 : Algebraic Functions

$a\ \mho \ b = a(b+1)-3$ $\displaystyle a\ \mho \ b = a(b+1)-3$

Quantity A Quantity B

$1\ \mho\ 1$ $\displaystyle 1\ \mho\ 1$ $2\ \mho\ 0$ $\displaystyle 2\ \mho\ 0$

Possible Answers:

The relationship cannot be determined from the information given.

Quantity A is greater

Quantity B is greater

Quantity A and Quantity B are equal

Correct answer:

Quantity A and Quantity B are equal

Explanation:

Since $a\ \mho \ b = a(b+1)-3$ $\displaystyle a\ \mho \ b = a(b+1)-3$, then we have that

$1\ \mho \ 1=1(1+1)-3=-1$ $\displaystyle 1\ \mho \ 1=1(1+1)-3=-1$

and

$2\ \mho \ 0=2(0+1)-3=-1$ $\displaystyle 2\ \mho \ 0=2(0+1)-3=-1$.

Thus, the two quantities are equal.

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ACT Math : ACT Math

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #101 : Algebraic Functions

Example Question #12 : Algebraic Functions

Example Question #11 : Algebraic Functions

Example Question #12 : How To Find F(X)

Example Question #342 : Algebra

Example Question #21 : How To Find F(X)

Example Question #22 : How To Find F(X)

Example Question #23 : How To Find F(X)

Example Question #24 : How To Find F(X)

Example Question #41 : Algebraic Functions

All ACT Math Resources

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