Question

b. $f(x)=2x^{2}-1$ from $x_1 = 1$ to $x_2=\frac{3}{4}$
net change:
average rate of change:

Explanation:

Step1: Recall net - change formula

Net change of \(y = f(x)\) from \(x = x_1\) to \(x = x_2\) is \(f(x_2)-f(x_1)\). Here, \(f(x)=2x^{2}-1\), \(x_1 = 1\), \(x_2=\frac{3}{4}\). First, find \(f(x_1)\) and \(f(x_2)\).
\[f(x_1)=2(1)^{2}-1=2 - 1=1\]
\[f(x_2)=2(\frac{3}{4})^{2}-1=2\times\frac{9}{16}-1=\frac{9}{8}-1=\frac{9 - 8}{8}=\frac{1}{8}\]

Step2: Calculate net - change

Net change \(=f(x_2)-f(x_1)=\frac{1}{8}-1=\frac{1 - 8}{8}=-\frac{7}{8}\)

Step3: Recall average - rate - of - change formula

The average rate of change of \(y = f(x)\) from \(x = x_1\) to \(x = x_2\) is \(\frac{f(x_2)-f(x_1)}{x_2 - x_1}\). We know \(f(x_2)-f(x_1)=-\frac{7}{8}\), \(x_2 - x_1=\frac{3}{4}-1=\frac{3 - 4}{4}=-\frac{1}{4}\)

Step4: Calculate average rate of change

Average rate of change \(=\frac{-\frac{7}{8}}{-\frac{1}{4}}=\frac{7}{8}\times4=\frac{7}{2}\)

Answer:

Net change: \(-\frac{7}{8}\); Average rate of change: \(\frac{7}{2}\)