Question

select the correct answer.
consider the functions f and g in the tables below.
\\( f(x) = 90x^2 + 180x + 92 \\)

x	y
1	362
2	812
3	1,442
4	2,252
5	3,242

\\( g(x) = 6^x \\)

x	y
1	6
2	36
3	216
4	1,296
5	7,776

which of the following statements is true?
a. at approximately \\( x = 4.39 \\), the rate of change of \\( f \\) is equal to the rate of change of \\( g \\).
b. as \\( x \\) increases, the rate of change of \\( f \\) exceeds the rate of change of \\( g \\).
c. as \\( x \\) increases, the rate of change of \\( g \\) exceeds the rate of change of \\( f \\).
d. for every value of \\( x \\), the rate of change of \\( g \\) exceeds the rate of change of \\( f \\).

Explanation:

Step1: Analyze function types

$f(x)$ is a quadratic function ($f(x) = 90x^2 + 180x + 92$), so its rate of change (derivative) is linear: $f'(x)=180x + 180$. $g(x)$ is an exponential function ($g(x)=6^x$), so its rate of change (derivative) is exponential: $g'(x)=6^x\ln(6)$.

Step2: Analyze rate of change behavior

- For quadratic $f(x)$, the rate of change $f'(x)$ is linear (increases at a constant rate).
- For exponential $g(x)$, the rate of change $g'(x)$ is exponential (increases at an increasing rate).

Step3: Evaluate options

- **Option A**: Solve $180x + 180 = 6^x\ln(6)$. Let's approximate. At $x = 4$, $f'(4)=180(4)+180 = 900$; $g'(4)=6^4\ln(6)=1296\ln(6)\approx1296\times1.7918\approx2323$. At $x = 3$, $f'(3)=180(3)+180 = 720$; $g'(3)=6^3\ln(6)=216\ln(6)\approx387$. Wait, earlier miscalculation. Wait, $\ln(6)\approx1.7918$. So $g'(3)=216\times1.7918\approx387$, $f'(3)=720$. At $x = 4$, $f'(4)=900$, $g'(4)=1296\times1.7918\approx2323$. Wait, maybe my initial derivative for $f(x)$: $f(x)=90x^2+180x + 92$, so $f'(x)=180x + 180$. Correct. For $g(x)=6^x$, $g'(x)=6^x\ln(6)$. Let's solve $180x + 180 = 6^x\ln(6)$. Let's try $x = 4.39$: $f'(4.39)=180(4.39)+180=180(5.39)\approx970.2$. $g'(4.39)=6^{4.39}\ln(6)$. $6^{4}=1296$, $6^{0.39}\approx6^{0.4}\approx2.0488$. So $6^{4.39}\approx1296\times2.0488\approx2655$. Then $g'(4.39)\approx2655\times1.7918\approx4757$. Wait, maybe my approximation is off. Wait, maybe I made a mistake. Wait, let's use a better approach. The linear function $f'(x)$ and exponential $g'(x)$. At some point, the exponential will overtake the linear, but when? Wait, actually, exponential functions grow faster than linear functions in the long run, but maybe at some point the linear is bigger first? Wait, no: $f'(x)$ is linear (slope 180), $g'(x)$ is exponential. Let's check at $x = 0$: $f'(0)=180$, $g'(0)=6^0\ln(6)=\ln(6)\approx1.79$. At $x = 1$: $f'(1)=360$, $g'(1)=6\ln(6)\approx10.75$. At $x = 2$: $f'(2)=540$, $g'(2)=36\ln(6)\approx64.5$. At $x = 3$: $f'(3)=720$, $g'(3)=216\ln(6)\approx387$. At $x = 4$: $f'(4)=900$, $g'(4)=1296\ln(6)\approx2323$. Wait, so at $x = 3$, $f'(3)=720 > g'(3)\approx387$. At $x = 4$, $f'(4)=900 < g'(4)\approx2323$. So there is a point where they cross. Wait, so solving $180x + 180 = 6^x\ln(6)$. Let's use trial. At $x = 3.5$: $f'(3.5)=180(3.5)+180=180(4.5)=810$. $g'(3.5)=6^{3.5}\ln(6)=6^3\times6^{0.5}\ln(6)=216\times\sqrt{6}\ln(6)\approx216\times2.4495\times1.7918\approx216\times4.39\approx948$. So $f'(3.5)=810 < g'(3.5)\approx948$? Wait, no, 216*2.4495≈529, 529*1.7918≈948. So at $x = 3.5$, $g'(x)\approx948$, $f'(x)=810$. At $x = 3$, $f'(3)=720$, $g'(3)\approx387$. So between $x = 3$ and $x = 3.5$, $g'(x)$ overtakes $f'(x)$? Wait, maybe my initial derivative for $f(x)$ is wrong? Wait, $f(x)=90x^2 + 180x + 92$. The difference between $f(1)-f(0)=362 - 92 = 270$; $f(2)-f(1)=812 - 362 = 450$; $f(3)-f(2)=1442 - 812 = 630$; $f(4)-f(3)=2252 - 1442 = 810$; $f(5)-f(4)=3242 - 2252 = 990$. So the average rate of change (which for quadratic is the same as the derivative at the midpoint) between $x=0$ and $x=1$ is 270 (midpoint $x=0.5$: $f'(0.5)=180(0.5)+180=270$). Between $x=1$ and $x=2$: 450 (midpoint $x=1.5$: $f'(1.5)=180(1.5)+180=450$). So the rate of change (average) for $f(x)$ between $x$ and $x+1$ is $f'(x + 0.5)=180(x + 0.5)+180=180x + 270$. Wait, maybe I confused instantaneous rate of change (derivative) with average rate of change. The problem says "rate of change"—probably instantaneous. So $f'(x)=180x + 180$, $g'(x)=6^x\ln(6)$. Let's solve $180x + 180 = 6^x\ln(6)$. Let's use $x = 4.39$: $f'(4.39)=180*4.39 + 180 ≈ 180*(5.39) ≈ 970.2$. $g'(4.39)=6^{4.39}\ln(6)$. Let's compute $6^{4.39}$: $\ln(6^{4.39})=4.39\ln(6)≈4.39*1.7918≈7.866$. So $6^{4.39}=e^{7.866}≈2650$. Then $g'(4.39)=2650*1.7918≈4748$. Wait, that's not equal. Maybe my approach is wrong. Alternatively, maybe the problem is about average rate of change (slope between points). Let's check average rate of change for $f(x)$: between $x=4$ and $x=5$: $\frac{3242 - 2252}{5 - 4}=990$. For $g(x)$ between $x=4$ and $x=5$: $\frac{7776 - 1296}{5 - 4}=6480$. Wait, no, earlier table for $f(x)$: at $x=4$, $y=2252$; $x=5$, $y=3242$. So difference 990. For $g(x)$: $x=4$, $y=1296$; $x=5$, $y=7776$. Difference 6480. So average rate of change for $f(x)$ between $x=4$ and $x=5$ is 990; for $g(x)$ it's 6480. Wait, but at $x=3$ to $x=4$: $f(x)$ difference: $2252 - 1442 = 810$; $g(x)$ difference: $1296 - 216 = 1080$. So at $x=3$ to $x=4$, $g(x)$ average rate (1080) exceeds $f(x)$ (810). At $x=2$ to $x=3$: $f(x)$ difference: $1442 - 812 = 630$; $g(x)$ difference: $216 - 36 = 180$. So $f(x)$ rate (630) exceeds $g(x)$ (180). At $x=3$ to $x=4$, $g(x)$ rate (1080) exceeds $f(x)$ (810). At $x=4$ to $x=5$, $g(x)$ rate (6480) exceeds $f(x)$ (990). So there is a point between $x=3$ and $x=4$ where the average rate of change of $g(x)$ overtakes $f(x)$. Wait, but the options:

- **Option A**: "At approximately x = 4.39, the rate of change of f is equal to the rate of change of g." Let's check instantaneous rate. $f'(x)=180x + 180$, $g'(x)=6^x\ln(6)$. Let's solve $180x + 180 = 6^x\ln(6)$. Let's use $x = 4$: $f'(4)=900$, $g'(4)=1296\ln(6)\approx2323$. $x = 3$: $f'(3)=720$, $g'(3)=216\ln(6)\approx387$. Wait, so at $x=3$, $f'(x) > g'(x)$; at $x=4$, $f'(x) < g'(x)$. So the crossing is between $x=3$ and $x=4$, not 4.39. Maybe the problem uses average rate of change. Let's check average rate between $x$ and $x+1$ for $f(x)$: $\Delta f = f(x+1) - f(x)$. For $x=4$: $\Delta f = 2252 - 1442 = 810$? Wait no, earlier table: $x=4$, $y=2252$; $x=5$, $y=3242$. So $\Delta f$ from $x=4$ to $x=5$ is 990. Wait, my mistake earlier: $f(3)=1442$, $f(4)=2252$ (difference 810), $f(4)=2252$, $f(5)=3242$ (difference 990). For $g(x)$: $g(3)=216$, $g(4)=1296$ (difference 1080), $g(4)=1296$, $g(5)=7776$ (difference 6480). So average rate of change (slope) for $f(x)$ between $x$ and $x+1$: at $x=3$: 810, $x=4$: 990. For $g(x)$: at $x=3$: 1080, $x=4$: 6480. So at $x=3$, $g(x)$ average rate (1080) > $f(x)$ (810). Wait, no, at $x=3$ (between 3 and 4), $g(x)$ average rate is 1080, $f(x)$ is 810. At $x=2$ (between 2 and 3): $f(x)$ average rate: 1442 - 812 = 630; $g(x)$: 216 - 36 = 180. So $f(x)$ rate > $g(x)$ rate. At $x=3$: $g(x)$ rate > $f(x)$ rate. So the crossing is between $x=2$ and $x=3$? Wait, no, at $x=2$ (between 2 and 3), $f$ rate 630, $g$ rate 180 (f > g). At $x=3$ (between 3 and 4), $f$ rate 810, $g$ rate 1080 (g > f). So crossing between $x=2$ and $x=3$? But option A says 4.39. Maybe the problem is about instantaneous rate. Let's recast:

- **Option B**: "As x increases, the rate of change of f exceeds the rate of change of g." But exponential rate of change (derivative) grows faster than linear, so eventually g's rate will exceed f's. So B is false.
- **Option C**: "As x increases, the rate of change of g exceeds the rate of change of f." Exponential derivative (exponential) grows faster than linear derivative (linear), so eventually (as x increases) g's rate will exceed f's. This is true in the long run.
- **Option D**: "For every value of x, the rate of change of g exceeds the rate of change of f." At x=0, f'(0)=180, g'(0)=ln(6)≈1.79. So f's rate > g's rate. So D is false.
- **Option A**: Let's check the equation $180x + 180 = 6^x\ln(6)$. Let's use x=4.39: f'(4.39)=180*4.39 + 180≈180*5.39≈970.2. g'(4.39)=6^4.39*ln(6). 6^4=1296, 6^0.39≈e^(0.39*ln6)≈e^(0.39*1.7918)≈e^0.70≈2.01. So 6^4.39≈1296*2.01≈2605. 2605*ln6≈2605*1.7918≈4668. Not equal to 970.2. So A is false.

Wait, maybe I made a mistake in the derivative of f(x). Wait, f(x)=90x² + 180x + 92. The derivative is f’(x)=180x + 180. Correct. g(x)=6^x, derivative is 6^x ln6. Correct.

So the correct option is C: As x increases, the rate of change of g exceeds the rate of change of f. Because exponential growth (rate of change) eventually outpaces linear growth (rate of change). At first, f's rate is higher (e.g., x=0,1,2,3), but as x increases, g's exponential rate of change (which grows exponentially) will overtake and surpass f's linear rate of change (which grows linearly).

Answer:

C. As x increases, the rate of change of g exceeds the rate of change of f.