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SAT II Math I : Solving Functions

Study concepts, example questions & explanations for SAT II Math I

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All SAT II Math I Resources

6 Diagnostic Tests 113 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #10 : Quadratic Inequalities

Which value for would satisfy the inequality $\small \small x^2 -5x -36< 0$ ?

Possible Answers:

$\small x = -5$

$\small x = 10$

$\small x = 1$

$\small x = -4$

Not enough information to solve

Correct answer:

$\small x = 1$

Explanation:

First, we can factor the quadratic to give us a better understanding of its graph. Factoring gives us: $\small (x+4)(x-9)< 0$ . Now we know that the quadratic has zeros at and . Furthermore this information reveals that the quadratic is positive. Using this information, we can sketch a graph like this:

Sketch inequality

We can see that the parabola is below the x-axis (in other words, less than ) between these two zeros and .

The only x-value satisfying the inequality $\small \small -4 < x < 9$ is $\small x=1$ .

The value of $\small x=-4$ would work if the inequality were inclusive, but since it is strictly less than instead of less than or equal to , that value will not work.

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Example Question #1 : Solving Other Functions

Simplify:

$\sqrt[3]{\sqrt[4]{x^{6}}}$

You may assume that is a nonnegative real number.

Possible Answers:

$\sqrt[7]{x^{6}}$

$\sqrt{x}$

$x^{2}$

$\frac{1}{x^{6}}$

$\frac{1}{x}$

Correct answer:

$\sqrt{x}$

Explanation:

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.

$\sqrt[3]{\sqrt[4]{x^{6}}} = \sqrt[3]{ \left (x^{6} \right ) ^{\frac{1}{4}}} =\left [ \left (x^{6} \right ) ^{\frac{1}{4} \right ] ^ {\frac{1}{3}}}$

Multiply the exponents, per the power of a power rule:

$\left [ \left (x^{6} \right ) ^{\frac{1}{4} \right ] ^ {\frac{1}{3}} = x ^{6 \cdot \frac{1}{4} \cdot \frac{1}{3}} = x^{\frac{1}{2}} = \sqrt{x}$

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Example Question #1 : Solving Other Functions

Define functions $p(x) = x^{5} + 2x^{2} - 7$ and $q(x)= x ^{4} - 3x$ .

p(c) = q(c) for exactly one value of on the interval (1.0, 1.5) .

Which of the following statements is correct about ?

Possible Answers:

$c \in ( 1.2, 1.3)$

$c \in ( 1.4, 1.5)$

$c \in (1.1, 1.2)$

$c \in ( 1.3, 1.4)$

$c \in (1.0, 1.1 )$

Correct answer:

$c \in ( 1.2, 1.3)$

Explanation:

Define

g(x) = p(x) - q(x)

$= (x^{5} + 2x^{2} - 7) - (x ^{4} - 3x )$

$= x^{5} - x ^{4}+ 2x^{2}+ 3x - 7$

Then if g(c) = 0 ,

it follows that

p(c) - q(c) = 0 ,

or, equivalently,

p(c) = q(c) .

By the Intermediate Value Theorem (IVT), if g(x) is a continuous function, and g(a) and g(b) are of unlike sign, then g(c) = 0 for some $c \in (a, b)$ . As a polynomial, g(x) is a continuous function, so the IVT applies here.

Evaluate g(x) for each of the following values: $\left \{ 1.0, 1.1, 1.2, 1.3, 1.4, 1.5\right \}$ :

$g(1.0)= 1.0^{5} - 1.0 ^{4}+ 2 \cdot 1.0^{2}+ 3 \cdot 1.0 - 7 \approx -2$

$g(1.1)= 1.1^{5} - 1.1 ^{4}+ 2 \cdot 1.1^{2}+ 3 \cdot 1.1 - 7 \approx - 1.13$

$g(1.2)= 1.2^{5} - 1.2 ^{4}+ 2 \cdot 1.2^{2}+ 3 \cdot 1.2 - 7 \approx -0.11$

$g(1.3)= 1.3^{5} - 1.3 ^{4}+ 2 \cdot 1.3^{2}+ 3 \cdot 1.3 - 7 \approx 1.4$

$g(1.4)= 1.4^{5} - 1.4 ^{4}+ 2 \cdot 1.4^{2}+ 3 \cdot 1.4 - 7 \approx 2.66$

$g(1.5)= 1.5^{5} - 1.5 ^{4}+ 2 \cdot 1.5^{2}+ 3 \cdot 1.5 - 7 \approx 4.53$

Only in the case of ( 1.2, 1.3) does it hold that g (x) assumes a different sign at both endpoints - g(1.2) < 0 < g(1.3) . By the IVT, g(c) = 0 , and p(c) = q(c) , for some $c \in (1.1, 1.2)$ .

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SAT II Math I : Solving Functions

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #10 : Quadratic Inequalities

Example Question #1 : Solving Other Functions

Example Question #1 : Solving Other Functions

All SAT II Math I Resources

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