SAT II Math I : Solving Other Functions

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

varsity tutors app store varsity tutors android store

Example Questions

Example Question #1 : Solving Other Functions

Simplify:

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

You may assume that \displaystyle x is a nonnegative real number.

Possible Answers:

\displaystyle \frac{1}{x}

\displaystyle x^{2}

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

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

\displaystyle \sqrt{x}

Correct answer:

\displaystyle \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.

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

Example Question #1 : Solving Other Functions

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

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

Which of the following statements is correct about \displaystyle c?

Possible Answers:

\displaystyle c \in ( 1.4, 1.5)

\displaystyle c \in ( 1.3, 1.4)

\displaystyle c \in ( 1.2, 1.3)

\displaystyle c \in (1.0, 1.1 )

\displaystyle c \in (1.1, 1.2)

Correct answer:

\displaystyle c \in ( 1.2, 1.3)

Explanation:

Define 

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

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

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

Then if \displaystyle g(c) = 0,

it follows that

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

or, equivalently,

\displaystyle p(c) = q(c).

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

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

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

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

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

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

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

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

Learning Tools by Varsity Tutors