Solving Functions - SAT Subject Test in Math II

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Question

Simplify:

$\sqrt[3]{\sqrt[2]{x^{-9}}}$

You may assume that is a nonnegative real number.

Answer

The best way to simplify a radical within a radical is to rewrite each root as a fractional exponent, then convert back.

First, rewrite the roots as exponents.

$\sqrt[3]{\sqrt[2]{x^{-9}}} = \sqrt[3]{\left ( x^{-9} \right ) ^{\frac{1}{2}} } = \left ( \left ( x^{-9} \right ) ^{\frac{1}{2}} \right )^{\frac{1}{3}} = x ^{-9 \cdot \frac{1}{2} \cdot \frac{1}{3}} } } = x ^{- \frac{3}{2} } }= \frac{1}{x^{\frac{3}{2}}}$

Then convert back to a radical and rationalizing the denominator:

$\frac{1}{x^{\frac{3}{2}}} = \frac{1}{\sqrt {x^{3}}} = \frac{1 \cdot \sqrt{x}}{\sqrt {x^{3} \cdot \sqrt{x}}} = \frac{ \sqrt{x}}{\sqrt {x^{4} }} =\frac{ \sqrt{x}}{x^{2}}$

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Question

Rewrite as a single logarithmic expression:

$\ln (y+ 1) - \ln (y + 2) + \ln (y + 3)$

Answer

Using the properties of logarithms

$\ln a - \ln b = \ln \frac{a}{b}$ and $\ln a + \ln b = \ln ab$ ,

simplify as follows:

$\ln (y+ 1) - \ln (y + 2) + \ln (y + 3)$

$= \ln \frac{y+ 1}{y + 2} + \ln (y + 3)$

$= \ln \left [\frac{y+ 1}{y + 2} \cdot (y + 3) \right ]$

$= \ln \frac{y^{2}+4y + 3}{y + 2}$

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Question

Simplify by rationalizing the denominator:

$\frac{11}{6- \sqrt{3}}$

Answer

Multiply the numerator and the denominator by the conjugate of the denominator, which is $6 + \sqrt{3}$ . Then take advantage of the distributive properties and the difference of squares pattern:

$\frac{11}{6- \sqrt{3}}$

$= \frac{11(6+ \sqrt{3})}{\left (6- \sqrt{3} \right )(6+ \sqrt{3})}$

$= \frac{11(6+ \sqrt{3})}{ 6^{2}- \left (\sqrt{3} \right ) ^{2}\right }$

$= \frac{11(6+ \sqrt{3})}{36-3 }$

$= \frac{11(6+ \sqrt{3})}{3 3 }$

$= \frac{6+ \sqrt{3}}{3 }$

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Question

Let $f(x) = 2^{x}+ 3^{2x}$ . What is the value of ?

Answer

Replace the integer as .

$f(-2) = 2^{-2}+ 3^{2(-2)} = 2^{-2}+ 3^{-4}$

Evaluate each negative exponent.

$2^{-2} = \frac{1}{2^2} = \frac{1}{4}$

$3^{-4} = \frac{1}{3^4} = \frac{1}{81}$

Sum the fractions.

$\frac{1}{4}+\frac{1}{81} = \frac{1(81)}{4(81)}+\frac{1(4)}{81(4)} = \frac{85}{324}$

The answer is: $\frac{85}{324}$

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Question

Find : $\sqrt{-3x-5} = 7$

Answer

Square both sides to eliminate the radical.

$(\sqrt{-3x-5}) ^2= 7^2$

Add five on both sides.

Divide by negative three on both sides.

$\frac{-3x}{-3} = \frac{54}{-3}$

The answer is:

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Question

If , what must be?

Answer

Replace the value of negative two with the x-variable.

There is no need to use the FOIL method to expand the binomial.

The answer is:

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Question

Let . What is the value of $f(\frac{7}{8})$ ?

Answer

Substitute the fraction as .

$f(\frac{7}{8}) = -3(\frac{7}{8})+9$

Multiply the whole number with the numerator.

$\frac{-21}{8}+9$

Convert the expression so that both terms have similar denominators.

$-\frac{21}{8}+\frac{9(8)}{1(8)} = -\frac{21}{8}+\frac{72}{8}$

The answer is: $\frac{51}{8}$

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Question

If , what must be?

Answer

A function of x equals five. This can be translated to:

This means that every point on the x-axis has a y value of five.

Therefore, .

The answer is:

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Question

Define and as follows:

$f(x) = \left{\begin{matrix} x+ 2,& x < 0 \ 4-x, & x \geq 0 \end{matrix}\right.$

$g(x) = \left{\begin{matrix} 4x ,& x < 3 \ 3x, & x \geq 3 \end{matrix}\right.$

Evaluate $(f \circ g) \left ( 3\right )$ .

Answer

$(f \circ g) \left ( x\right ) = f (g(x))$ by definition.

$(f \circ g) \left ( 3\right ) = f (g (3))$

on the set $[3, \infty)$ , so

$g(3) = 3 \cdot 3 = 9$ .

on the set $[0 , \infty)$ , so

.

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Question

Define functions and as follows:

$g(x) = \left{\begin{matrix} x+ 6 &x< 0 \ x- 3 & x \ge 0 \end{matrix}\right.$

Evaluate $(f \circ g) \left ( 2\frac{1}{3}\right )$ .

Answer

$(f \circ g) \left ( 2\frac{1}{3}\right )=f\left ( g \left ( 2\frac{1}{3}\right ) \right )$

First, we evaluate $g \left ( 2\frac{1}{3}\right )$ . Since $2\frac{1}{3} \ge 0$ , the definition of for $x \ge 0$ is used, and

$g \left ( 2\frac{1}{3}\right ) = 2\frac{1}{3} - 3 = -\frac{2}{3}$

Since

, then

$(f \circ g) \left ( 2\frac{1}{3}\right )$

$=f\left ( g \left ( 2\frac{1}{3}\right ) \right )$

$=f\left ( -\frac{2}{3} \right )$

$= 4 \left ( -\frac{2}{3} \right )+7$

$= -2\frac{2}{3} +7$

$= 4\frac{1}{3}$

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Question

Define functions and as follows:

$f(x) = \left{\begin{matrix} 2x+ 1 &x< 0 \ 3x & x \ge 0 \end{matrix}\right.$

$g(x) = \left{\begin{matrix} x+ 9 &x< 0 \ x- 2 & x \ge 0 \end{matrix}\right.$

Evaluate $(f \circ g) \left (-1.7\right )$ .

Answer

$(f \circ g) \left ( -1.7\right )=f\left ( g \left ( -1.7\right ) \right )$

First we evaluate . Since , we use the definition of for the values in the range :

Therefore,

$(f \circ g) \left ( -1.7\right )=f\left ( g \left ( -1.7\right ) \right ) = f (7.3)$

Since $7.3 \ge 0$ , we use the definition of for the range $x \ge 0$ :

$f (7.3) = 3 \cdot 7.3 = 21.9$

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Question

Define function as follows:

$f(x) = \left{\begin{matrix} 2x- 5 &x< 0 \ -3x & x \ge 0 \end{matrix}\right.$

Give the range of .

Answer

The range of a piecewise function is the union of the ranges of the individual pieces, so we examine both of these pieces.

If , then . To find the range of on the interval $(-\infty, 0)$ , we note:

$2x < 2 \cdot 0$

The range of this portion of is $(-\infty, -5)$ .

If $x \ge 0$ , then . To find the range of on the interval $[0, \infty)$ , we note:

$x \ge 0$

$-3x \le -3 \cdot 0$

$-3x \le 0$

$f(x)\le 0$

The range of this portion of is $(-\infty, 0 ]$

The union of these two sets is $(-\infty, 0 ]$ , so this is the range of over its entire domain.

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Question

Define function as follows:

$f(x) = \left{\begin{matrix} x- 7 &x< -1 \ 6-x & x > 1 \end{matrix}\right.$

Give the range of .

Answer

The range of a piecewise function is the union of the ranges of the individual pieces, so we examine both of these pieces.

If , then .

To find the range of on the interval $(-\infty, -1)$ , we note:

The range of on $(-\infty, -1)$ is $(-\infty, -8)$ .

If , then .

To find the range of on the interval $(1, \infty )$ , we note:

The range of on $(1, \infty )$ is $(-\infty, 5)$ .

The range of on its entire domain is the union of these sets, or $(-\infty, 5)$ .

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Question

Define functions and as follows:

$f(x) = \left{\begin{matrix} 2x &x< -1 \ 3x & x > 1 \end{matrix}\right.$

$g(x) = \left{\begin{matrix} x+ 5 &x< -1 \ x- 2 & x > 1 \end{matrix}\right.$

Evaluate Evaluate $(f \circ g)(3)$ .

Answer

$(f \circ g)(3) = f(g(3))$

First, evaluate using the definition of for :

Therefore,

$(f \circ g)(3) = f(g(3)) = f(1)$

However, is not in the domain of .

Therefore, $(f \circ g)(3)$ is an undefined quantity.

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Question

Define functions and as follows:

$f(x) = \left{\begin{matrix} 2x &x< -1 \ 3x & x > 1 \end{matrix}\right.$

$g(x) = \left{\begin{matrix} x+ 5 &x< -1 \ x- 2 & x > 1 \end{matrix}\right.$

Evaluate $(g \circ f)(-3)$ .

Answer

$(g \circ f)(-3)= g(f(-3))$

First, evaluate using the definition of for :

Therefore,

$(g \circ f)(-3)= g(f(-3)) = g(-6)$

Evaluate using the definition of for :

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Question

Which of the following would be a valid alternative definition for the provided function?

$f(x) = \left | \frac{3}{4} x - \frac{7}{3} \right |$

Answer

The absolute value of an expression is defined as follows:

for $a \ge 0$

for

Therefore,

$f(x) = \left | \frac{3}{4} x - \frac{7}{3} \right | = \frac{3}{4} x - \frac{7}{3}$

if and only if

$\frac{3}{4} x - \frac{7}{3} \ge 0$ .

Solving this condition for :

$\frac{3}{4} x - \frac{7}{3} + \frac{7}{3} \ge 0 + \frac{7}{3}$

$\frac{3}{4} x \ge \frac{7}{3}$

$\frac{4}{3} \cdot \frac{3}{4} x \ge \frac{4}{3} \cdot \frac{7}{3}$

$x \ge \frac{28}{9}$

Therefore, $f(x) = \frac{3}{4} x - \frac{7}{3}$ for $x \ge \frac{28}{9}$ .

Similarly,

$f(x) = - \left (\frac{3}{4} x - \frac{7}{3} \right ) = - \frac{3}{4} x + \frac{7}{3}$ for $x < \frac{28}{9}$ .

The correct response is therefore

$f(x) = \left{\begin{matrix} - \frac{3}{4} x + \frac{7}{3} & \textrm{for }x < \frac{28}{9}\ \ \frac{3}{4} x - \frac{7}{3}& \textrm{for }x \ge \frac{28}{9} \end{matrix}\right.$

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Question

Define two functions as follows:

$f(x) = \begin{Bmatrix} 2x- 7& \textrm{for } x < 0 \ & \ 3x+ 3& \textrm{for } x\ge 0 \end{matrix}$

$g(x) = \begin{Bmatrix} 3x+ 3 & \textrm{for } x < 0 \ & \ 2x- 7 & \textrm{for } x\ge 0 \end{matrix}$

Evaluate $(f \circ g) (5)$ .

Answer

By definition, $(f \circ g) (5) = f \left [ g (5) \right ]$

First, evaluate , using the definition of for nonnegative values of . Substituting for 5:

$g(5) = 2\cdot 5 -7= 10 -7 = 3$

$(f \circ g) (5) = f \left [ g (5) \right ] = f(3)$ ; evaluate this using the definition of for nonnegative values of :

$f(3) = 3 \cdot 3+ 3 = 9 + 3 = 12$

12 is the correct value.

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Question

A baseball is thrown straight up with an initial speed of 60 feet per second by a man standing on the roof of a 100-foot high building. The height of the baseball in feet as a function of time in seconds is modeled by the function

$h(t) = -16t^{2}+60t + 100$

To the nearest tenth of a second, how long does it take for the baseball to hit the ground?

Answer

When the baseball hits the ground, the height is 0, so we set $h\left ( t\right ) = 0$ . and solve for .

$-16t^{2}+60t + 100 = 0$

This can be done using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

Set :

$t = \frac{-60 \pm \sqrt{60^{2}-4(-16)100}}{2(-16)}$

$t = \frac{-60 \pm \sqrt{3,600- (- 6,400)}}{-32}$

$t = \frac{-60 \pm \sqrt{3,600+6,400}}{-32}$

$t = \frac{-60 \pm \sqrt{10,000}}{-32}$

$t = \frac{-60 \pm 100}{-32}$

One possible solution:

$t = \frac{-60 + 100}{-32} = \frac{40}{-32} = -1.25$

We throw this out, since time must be positive.

The other:

$t = \frac{-60 - 100}{-32} = \frac{-160}{-32} = 5$

This solution, we keep. The baseball hits the ground in 5 seconds.

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Question

If $f(x) = 2\sqrt{x}-\sqrt{x-3}$ , what is the value of ?

Answer

Substitute the value of negative three as .

$f(-3) = 2\sqrt{-3}-\sqrt{(-3)-3}$

$2\sqrt{-3}-\sqrt{-6} = 2\sqrt{-1}\cdot \sqrt{3} - \sqrt{-1}\cdot \sqrt{6}$

The terms will be imaginary. We can factor out an $i = \sqrt{-1}$ out of the right side. Replace them with .

The answer is: $2i\sqrt3 -i\sqrt6$

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Question

Give the period of the graph of the equation

$f(x) = \frac{2}{9} \sin 8 x$

Answer

The period of the graph of a sine function $f(x)= A \sin Bx$ is $\frac{2 \pi}{B}$ , or $2 \pi \div B$ .

Since ,

$2 \pi \div 8 = \frac{\pi }{4}$ .

This answer is not among the given choices.

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