Answer:
2/π
h(x) is concave up on the intervals (-∞, -1/√3) and (1/√3, ∞).
h(x) is concave down on the interval (-1/√3, 1/√3).
Step-by-step explanation:
f(x) = sin x
The average value of a function between x=a and x=b is:
avg = 1/(b−a) ∫ₐᵇ f(x) dx
avg = 1/(π−0) ∫₀ᵖⁱ sin x dx
avg = 1/π (-cos x) |₀ᵖⁱ
avg = 1/π (-cos π − (-cos 0))
avg = 1/π (1 + 1)
avg = 2/π
h(x) = x⁴/12 − x²/6
Find the second derivative.
h'(x) = x³/3 − x/3
h"(x) = x² − ⅓
Factor using difference of squares.
h"(x) = (x − 1/√3) (x + 1/√3)
h"(x) = 0 when x = ±1/√3. Evaluate the sign of h"(x) in each interval.
-∞ < x < -1/√3, h"(x) > 0.
-1/√3 < x < 1/√3, h"(x) < 0.
1/√3 < x < ∞, h"(x) > 0.
h(x) is concave up on the intervals (-∞, -1/√3) and (1/√3, ∞).
h(x) is concave down on the interval (-1/√3, 1/√3).
