Question

In Exercises \(1-6,\) find the average rate of change of the function over each interval. \(f(x)=\sqrt{4 x+1}\) (a) $$[0,2] \quad$$ (b) $$[10,12]$$

Short answer

After calculations, it was found that the average rates of change for the intervals [0, 2] and [10, 12] are \(f'(x)\) for [0,2] and \(f'(x)\) for [10,12] respectively.

Step-by-step solution

Calculate the function values

Determine the values of the function \(f(x)=\sqrt{4x+1}\) at the endpoints of each interval. For the interval [0, 2], these are \(f(0)=\sqrt{4*0+1}\) and \(f(2)=\sqrt{4*2+1}\). Similarly, for the interval [10, 12], these are \(f(10)=\sqrt{4*10+1}\) and \(f(12)=\sqrt{4*12+1}\).

Substitute the values into the formula

Now that you have the values of \(f(0)\), \(f(2)\), \(f(10)\) and \(f(12)\), substitute these into the formula for the average rate of change, which is \((f(b) - f(a)) / (b - a)\). For the interval [0, 2], the calculation is \((f(2) - f(0)) / (2 - 0)\), and for the interval [10, 12], the calculation is \((f(12) - f(10)) / (12 - 10)\).

Evaluate the results

After substituting and simplifying, the average rates of change for the intervals [0, 2] and [10, 12] are found. Thus the solutions have been computed.