(a) It looks like you're saying
[tex]\displaystyle F(x) = \int_0^x (t - 3t^2 + 7) \, dt[/tex]
Find the critical points of F(x). By the fundamental theorem of calculus,
F'(x) = x - 3x² + 7
The critical points are where the derivative vanishes. Using the quadratic formula,
x - 3x² + 7 = 0 ⇒ x = (1 ± √85)/6
Compute the second derivative of F :
F''(x) = 1 - 6x
Check the sign of the second derivative at each critical point.
• x = (1 + √85)/6 ≈ 1.703 ⇒ F''(x) < 0
• x = (1 - √85)/6 ≈ -1.370 ⇒ F''(x) > 0
This tells us F attains a minimum of
[tex]F\left(\dfrac{1-\sqrt{85}}6\right) \approx \boxed{-6.080}[/tex]
(b) Split up the domain of F at the critical points, and check the sign of F'(x) over each subinterval.
• over (-∞, -1.370), consider x = -2; then F'(x) = -7 < 0
• over (-1.370, 1.703), consider x = 0; then F'(x) = 7 > 0
• over (1.703, ∞), consider x = 2; then F'(x) = -3 < 0
This tells us that
• F(x) is increasing over ((1 - √85)/6, (1 + √85)/6)
• F(x) is decreasing over (-∞, (1 - √85)/6) and ((1 + √85)/6, ∞)
(c) Solve F''(x) = 0 to find the possible inflection points of F(x) :
F''(x) = 1 - 6x = 0 ⇒ 6x = 1 ⇒ x = 1/6
Split up the domain at the inflection point and check the sign of F''(x) over each subinterval.
• over (-∞, 1/6), consider x = 0; then F''(x) = 1 > 0
• over (1/6, ∞), consider x = 2; then F''(x) = -11 < 0
This tells us that
• F(x) is concave up over (-∞, 1/6)
• F(x) is concave down over (1/6, ∞)
