Questions: Calculate the average rate of change of the given function f over the intervals [a, a+h] where h=1, 0.1, 0.01, 0.001, and 0.0001. (Technology is recommended for the cases h=0.01, 0.001, and 0.0001.) (Round your answers to seven decimal places.) f(x) = x^2 / 6 ; a=1 h=1 h=0.1 h=0.01 h=0.001 h=0.0001

Calculate the average rate of change of the given function f over the intervals [a, a+h] where h=1, 0.1, 0.01, 0.001, and 0.0001. (Technology is recommended for the cases h=0.01, 0.001, and 0.0001.) (Round your answers to seven decimal places.)
f(x) = x^2 / 6 ; a=1
h=1
h=0.1
h=0.01
h=0.001
h=0.0001

Transcript text: Calculate the average rate of change of the given function $f$ over the intervals $[a, a+h]$ where $h=1,0.1,0.01,0.001$, and 0.0001 . (Technology is recommended for the cases $h=0.01,0.001$, and 0.0001 .) (Round your answers to seven decimal places.) \[ f(x)=\frac{x^{2}}{6} ; \quad a=1 \] \[ \begin{array}{l} h=1 \\ h=0.1 \\ h=0.01 \\ h=0.001 \\ h=0.0001 \end{array} \] $\square$ $\square$ $\square$ $\square$ $\square$

Solution

Solution Steps

To calculate the average rate of change of the function $ f(x) = \frac{x^2}{6} $ over the interval $[a, a+h]$, we use the formula for the average rate of change: $\frac{f(a+h) - f(a)}{h}$. We will compute this for each given value of $ h $ while rounding the results to seven decimal places.

Step 1: Define the Function and Interval

We are given the function $ f(x) = \frac{x^2}{6} $ and need to calculate the average rate of change over the interval $[a, a+h]$ where $ a = 1 $ and $ h $ takes on several values.

Step 2: Calculate the Average Rate of Change

The average rate of change of a function $ f $ over an interval $[a, a+h]$ is given by the formula: \[ \frac{f(a+h) - f(a)}{h} \] We will compute this for each specified value of $ h $.

Step 3: Compute for Each $ h $

For $ h = 1 $: \[ \frac{f(1+1) - f(1)}{1} = \frac{f(2) - f(1)}{1} = \frac{\frac{2^2}{6} - \frac{1^2}{6}}{1} = \frac{\frac{4}{6} - \frac{1}{6}}{1} = \frac{3}{6} = 0.5 \]
For $ h = 0.1 $: \[ \frac{f(1+0.1) - f(1)}{0.1} = \frac{f(1.1) - f(1)}{0.1} = \frac{\frac{1.1^2}{6} - \frac{1^2}{6}}{0.1} \approx 0.35 \]
For $ h = 0.01 $: \[ \frac{f(1+0.01) - f(1)}{0.01} = \frac{f(1.01) - f(1)}{0.01} = \frac{\frac{1.01^2}{6} - \frac{1^2}{6}}{0.01} \approx 0.335 \]
For $ h = 0.001 $: \[ \frac{f(1+0.001) - f(1)}{0.001} = \frac{f(1.001) - f(1)}{0.001} = \frac{\frac{1.001^2}{6} - \frac{1^2}{6}}{0.001} \approx 0.3335 \]
For $ h = 0.0001 $: \[ \frac{f(1+0.0001) - f(1)}{0.0001} = \frac{f(1.0001) - f(1)}{0.0001} = \frac{\frac{1.0001^2}{6} - \frac{1^2}{6}}{0.0001} \approx 0.33335 \]