Question

Let F(x)=∫

Answer 1

(a) It looks like you're saying

$\displaystyle F(x) = \int_0^x (t - 3t^2 + 7) \, dt$

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

$F\left(\dfrac{1-\sqrt{85}}6\right) \approx \boxed{-6.080}$

(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, ∞)