The function is defined by for . The function is an increasing linear function. In the -plane, the graphs of and intersect at exactly two points, and . When , which of the following must be true?

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Super-Hard SAT Math Question 119 | Free SAT Prep | Aniko

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Question 119·200 Super-Hard SAT Math Questions·Advanced Math

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The function $f$ is defined by $f(x)=\log_2 x$ for $x>0$ . The function $g$ is an increasing linear function. In the $xy$ -plane, the graphs of $y=f(x)$ and $y=g(x)$ intersect at exactly two points, $(2,1)$ and $(8,3)$ . аnikο‌.а‌i

When $g(x) > f(x)$ , which of the following must be true?

When two continuous graphs intersect at exactly two $x$ -values, split the number line into the three intervals determined by those intersection points. Then test one convenient $x$ -value in each interval to determine whether $g(x)-f(x)$ is positive or negative there; the sign cannot change within an interval without creating another intersection. A-n-і-k-ο.a‍і

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Find the line

Use the two intersection points to find the slope and equation of $g(x)$ . Рrοpеrty of Anі‍ko.аі

Split the x-axis into regions

The intersections happen at $x=2$ and $x=8$ , so consider the intervals $(0,2)$ , $(2,8)$ , and $(8,\infty)$ .

Test one x-value in each region

Pick simple values like $x=1$ , $x=4$ , and $x=16$ and compare $g(x)$ to $f(x)$ .

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Graph the two functions

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$y=\log(x)/\log(2)$
$y=x/3+1/3$

(Using $\log(x)/\log(2)$ graphs $\log_2 x$ .)

Verify the intersection x-values

Click the intersection points and confirm they occur at $x=2$ and $x=8$ .

Check where the line is above the curve

Look at the graph on the intervals $0<x<2$ , $2<x<8$ , and $x>8$ . Identify where the line lies above the logarithm curve; those x-values correspond to $g(x)>f(x)$ .

Step-by-step Explanation

Write the equation of the line $g$

Because $g$ passes through $(2,1)$ and $(8,3)$ , its slope is

\frac{3-1}{8-2}=\frac{2}{6}=\frac{1}{3}.

Using point-slope form with $(2,1)$ :

\begin{aligned} y-1 &= \frac{1}{3}(x-2) \\ y &= \frac{1}{3}x+\frac{1}{3} \end{aligned}

So $g(x)=\frac{1}{3}x+\frac{1}{3}$ .

Determine the sign on $(2,8)$

Test a value between $2$ and $8$ , such as $x=4$ .

f(4)=\log_2 4=2

g(4)=\frac{1}{3}(4)+\frac{1}{3}=\frac{5}{3}

Since $\frac{5}{3}<2$ , we have $g(4)<f(4)$ , so $g(x)-f(x)<0$ somewhere in $(2,8)$ .

Use “exactly two intersections” to get the full solution set

Since the graphs intersect only at $x=2$ and $x=8$ , the continuous function $g(x)-f(x)$ can’t change sign within any of these intervals: $(0,2)$ , $(2,8)$ , and $(8,\infty)$ .

From Step 2, $g(x)-f(x)<0$ at $x=4$ , so $g(x)<f(x)$ for all $2<x<8$ .
Test one value in $(0,2)$ , say $x=1$ :

g(1)=\frac{2}{3},\quad f(1)=\log_2 1=0,

so $g(1)>f(1)$ and therefore $g(x)>f(x)$ for all $0<x<2$ .

Test one value in $(8,\infty)$ , say $x=16$ :

g(16)=\frac{17}{3},\quad f(16)=\log_2 16=4,

so $g(16)>f(16)$ and therefore $g(x)>f(x)$ for all $x>8$ .

Thus, when $g(x)>f(x)$ , it must be true that $0<x<2$ or $x>8$ . аniko.⁠a‍і/ѕ⁠аt

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