Question

Let F(x)=∫

t−3

t

2+7

for − ∞ < x < ∞

x

(a) Find the value of x where F attains its minimum value.

(b) Find intervals over which F is only increasing or only decreasing.

(c) Find open intervals over which F is only concave up or only concave down.

Answer 1

$$F(x)=\displaystyle\int_{0}^x(t-3t^2+7)dt$$

By the Fundamental Theorem of Calculus

$$F'(x)=x-3x^2+7$$

Find the critical points of $F(x)$

$$F'(x)=0=>x-3x^2+7=0$$

$$3x^2-x-7=0$$

$$D=(-1)^2-4(3)(-7)=85$$

$$x=\dfrac{1\pm\sqrt{85}}{2(3)}$$

$$x_1=\dfrac{1-\sqrt{85}}{6}, x_2=\dfrac{1+\sqrt{85}}{6}$$

If $x<\dfrac{1-\sqrt{85}}{6}, F'(x)<0, F(x)$ decreases.

If $\dfrac{1-\sqrt{85}}{6}<x<\dfrac{1+\sqrt{85}}{6}, F'(x)>0, F(x)$ increases.

If $x>\dfrac{1+\sqrt{85}}{6}, F'(x)<0, F(x)$ decreases.

(a)The function $F(x)$ has the local minimum at $x=\dfrac{1-\sqrt{85}}{6}.$

(b) The function $F(x)$ increases on

$$(\dfrac{1-\sqrt{85}}{6}, \dfrac{1+\sqrt{85}}{6}).$$

The function $F(x)$ decreases on

$$(-\infin, \dfrac{1-\sqrt{85}}{6})\cup(\dfrac{1+\sqrt{85}}{6}, \infin)$$

(c)

$$F''(x)=(x-3x^2+7)'=1-6x$$

Find the inflection point(s)

$$F''(x)=0=>1-6x=0=>x=\dfrac{1}{6}$$

If $x<\dfrac{1}{6}, F''(x)>0, F(x)$ is concave up.

If $x>\dfrac{1}{6}, F''(x)<0, F(x)$ is concave down.

The function $F(x)$ is only concave up on $(-\infin, \dfrac{1}{6}).$

The function $F(x)$ is only concave down on $(\dfrac{1}{6}, \infin).$