Question

this shows two functions.
$f(x) = 2x^3 - 8x^2 + 5x + 2$
$g(x) = (1.5)^x$
what is the largest integer value of the domain for which $f(x)$ is greater than $g(x)$?
a. 0
b. 29
c. 3
d. 25

Explanation:

Step1: Test x=25

$f(25)=2(25)^3 -8(25)^2 +5(25)+2=2(15625)-8(625)+125+2=31250-5000+127=26377$
$g(25)=(1.5)^{25}\approx1002.55$
$f(25)>g(25)$

Step2: Test x=26

$f(26)=2(26)^3 -8(26)^2 +5(26)+2=2(17576)-8(676)+130+2=35152-5408+132=29876$
$g(26)=(1.5)^{26}\approx1503.83$
$f(26)>g(26)$? Wait, wait—wait, no, wait: Wait, $(1.5)^{25}\approx1002$, $(1.5)^{26}=1.5×1002≈1503$, still less than 29876. Wait, check x=29:
$f(29)=2(24389)-8(841)+145+2=48778-6728+147=42197$
$g(29)=(1.5)^{29}=(1.5)^{25}×(1.5)^4≈1002×5.0625≈5073$
Wait, no—wait, maybe I miscalculated. Wait, no, wait the options: B is 29, D is25. Wait, wait, let's check x=3 first:
$f(3)=2(27)-8(9)+15+2=54-72+17=-1$
$g(3)=3.375$, so f(3)<g(3). x=4: $f(4)=2(64)-8(16)+20+2=128-128+22=22$; $g(4)=5.0625$, f>g. x=10: $f(10)=2000-800+50+2=1252$; $g(10)≈57.67$, f>g. x=20: $f(20)=16000-3200+100+2=12902$; $g(20)≈(1.5)^{20}≈3325.26$, f>g. x=25: f=26377>1002. x=29: f=42197>g(29)≈5073? Wait, but why option D? Wait, maybe the original function was a typo? Wait no—the question says "largest integer value of the domain for which f(x)>g(x)". Wait, but exponential grows faster than polynomial eventually. Wait, when does $(1.5)^x$ overtake $2x^3-8x^2+5x+2$? Let's estimate: Let’s set $2x^3≈(1.5)^x$. Take log: $\ln2 +3\ln x≈x\ln1.5$. For x=30: $\ln2+3\ln30≈0.693+3×3.401≈0.693+10.203=10.896$; $30×0.405≈12.15>10.896$. So $(1.5)^{30}≈e^{12.15}≈188000$, while $f(30)=2×27000 -8×900 +150+2=54000-7200+152=46952<188000$. So x=29: $\ln2+3\ln29≈0.693+3×3.367≈0.693+10.101=10.794$; $29×0.405≈11.745>10.794$, so $(1.5)^{29}≈e^{11.745}≈127000$, f(29)=42197<127000. x=25: $\ln2+3\ln25≈0.693+3×3.2189≈0.693+9.6567=10.3497$; $25×0.405=10.125<10.3497$, so $(1.5)^{25}≈e^{10.125}≈24800$? Wait, no, earlier mistake: $\ln(1.5)≈0.4055$, so $x\ln1.5$ for x=25 is 25×0.4055=10.1375; $3\ln x +\ln2$ for x=25 is 3×3.2189+0.693≈10.3497>10.1375, so $2x^3>1.5^x$. For x=26: 26×0.4055=10.543; 3×ln26+ln2≈3×3.258+0.693≈10.467<10.543, so $1.5^{26}>2x^3$. So f(26)=2×26³-8×26²+5×26+2≈2×17576-8×676+130+2=35152-5408+132=29876; $1.5^{26}≈e^{10.543}≈38000>29876$. So f(26)<g(26). x=25: $1.5^{25}=e^{10.1375}≈24800$, f(25)=2×15625-8×625+125+2=31250-5000+127=26377>24800. So largest integer x is25.

Step3: Confirm options

x=25 is option D, and x=26 is f<g. So answer is D.

Answer:

D. 25