Question

Let f(x) = log2(4x + 8x ln 2). For all values of a for which the function is defined, what is the instantaneous rate of change at x = a?

StartFraction 1 Over a EndFraction
StartFraction 1 Over a l n 2 EndFraction
StartFraction 1 Over 2 a l n 2 EndFraction
StartFraction 1 Over 4 a (1 + 2 l n 2) EndFraction

Answer 1

The instantaneous rate of change of the function f(x) at any point x = a is given by the derivative evaluated at that point. The derivative of f(x), after applying the logarithmic differentiation rules and simplification, is 1/(4a(1 + 2 ln 2)).

Step-by-step explanation:

To find the instantaneous rate of change of the function f(x) = log2(4x + 8x ln 2) at any point x = a, where the function is defined, we need to calculate its derivative and then evaluate it at x = a. The derivative of a logarithmic function can be computed using the chain rule.

The general rule for the derivative of logb(x) is 1 / (x ln b). Applying this to our function, we get:

• The derivative of 4x is 4 and the derivative of 8x ln 2 is 8 ln 2, since ln 2 is a constant.
• Summing these derivatives due to the sum inside the log, we get a combined derivative of 4 + 8 ln 2.
• Now we apply the log derivative rule: f'(x) = (4 + 8 ln 2) / (4x + 8x ln 2 ln 2)
• Simplify to get f'(x) = 1 / (x(1 + 2 ln 2)).
• The instantaneous rate of change at x = a is therefore f'(a) = 1 / (a(1 + 2 ln 2)), which simplifies to the fourth answer choice, 1 / (4a(1 + 2 ln 2)).

By understanding the properties of logarithms and the chain rule, we have correctly identified the instantaneous rate of change for the function at any value a.