Question

which function has an average rate of change over the interval
x = 5 to x = 8 that is closest to the rate of change of f(x) over
the same interval?
a g(x)=0.8x² - 4
b h(x)=1/6x³
c k(x)=1.5^x + 3
d j(x)=35log(x)

Explanation:

Step1: Calculate the average rate of change of \(f(x)\)

The formula for the average rate of change of a function \(y = f(x)\) over the interval \([a,b]\) is \(\frac{f(b)-f(a)}{b - a}\).
Here, \(a = 5\), \(b = 8\), \(f(5)=12.6\), \(f(8)=30.6\)

$$ \frac{f(8)-f(5)}{8 - 5}=\frac{30.6-12.6}{3}=\frac{18}{3}=6 $$

Step2: Calculate the average rate of change for \(g(x)=0.8x^{2}-4\)

For \(g(x)=0.8x^{2}-4\), when \(x = 5\), \(g(5)=0.8\times5^{2}-4=0.8\times25 - 4=20 - 4 = 16\)
When \(x = 8\), \(g(8)=0.8\times8^{2}-4=0.8\times64-4=51.2 - 4 = 47.2\)

$$ \frac{g(8)-g(5)}{8 - 5}=\frac{47.2-16}{3}=\frac{31.2}{3}=10.4 $$

Step3: Calculate the average rate of change for \(h(x)=\frac{1}{6}x^{3}\)

When \(x = 5\), \(h(5)=\frac{1}{6}\times5^{3}=\frac{125}{6}\approx20.83\)
When \(x = 8\), \(h(8)=\frac{1}{6}\times8^{3}=\frac{512}{6}\approx85.33\)

$$ \frac{h(8)-h(5)}{8 - 5}=\frac{\frac{512}{6}-\frac{125}{6}}{3}=\frac{\frac{387}{6}}{3}=\frac{387}{18}=21.5 $$

Step4: Calculate the average rate of change for \(k(x)=1.5^{x}+3\)

When \(x = 5\), \(k(5)=1.5^{5}+3=(1.5)^{2}\times(1.5)^{2}\times1.5+3 = 2.25\times2.25\times1.5+3=7.59375+3 = 10.59375\)
When \(x = 8\), \(k(8)=1.5^{8}+3=(1.5^{4})^{2}+3=(5.0625)^{2}+3 = 25.62890625+3=28.62890625\)

$$ \frac{k(8)-k(5)}{8 - 5}=\frac{28.62890625 - 10.59375}{3}=\frac{18.03515625}{3}=6.01171875 $$

Step5: Calculate the average rate of change for \(j(x)=35\log(x)\)

When \(x = 5\), \(j(5)=35\log(5)\approx35\times0.69897 = 24.46395\)
When \(x = 8\), \(j(8)=35\log(8)=35\times0.90309 = 31.60815\)

$$ \frac{j(8)-j(5)}{8 - 5}=\frac{31.60815-24.46395}{3}=\frac{7.1442}{3}=2.3814 $$

Answer:

\(C\)