Let F(x)=∫x96ln(2t)dt, for x≥9.

Question

(a) F′(x)=6/(ln(2x))
(b) On what interval or intervals is F increasing?
x∈
(c) On what interval or intervals is the graph of F concave up?
x∈

Answer 1

(a) Applying the Fundamental Theorem of Calculus, we have F'(x) = 96ln(2x) ⋅ d/dx(x) = 96ln(2x)⋅1 = 96ln(2x)/1 = 6ln(2x)/ln(2) = 6/ln(2x).

(b) To determine the intervals on which F is increasing, we need to analyze the sign of F'(x). F'(x) is positive if 6/ln(2x) > 0 or if ln(2x) < 0, which is true if 0 < x < 1/2. F'(x) is negative if 6/ln(2x) < 0 or if ln(2x) > 0, which is true if x > 1/2. Therefore, F is increasing on the interval (0, 1/2) and decreasing on the interval (1/2, ∞).

(c) To determine the intervals on which the graph of F is concave up, we need to analyze the sign of F''(x). Using the chain rule, we can find that F''(x) = 6/(xln(2))^2. Since the denominator is always positive, we can determine the sign of F''(x) by analyzing the sign of the numerator. F''(x) is positive if 6/(xln(2))^2 > 0 or if x > 0, which means that the graph of F is concave up on the interval (0, ∞).