Calculus help, please?

Question

let f(x) = x^3+ax^2+ bx+c where a,b,and c are constants with a › or equal to zero, and bis greater then zero.

A) over what intervals is f concave up?

B) show that f must have exactly one inflection point.

C? given that 0,-2 is the inflection point of f, compute a and c and then show that f has no critical point.

Answer 1

A) Take the second derivative.
f'(x)=3x^2+2ax+b
f''(x)=6x+2a
Find the roots of the second derivative.
0=6x+2a
x=-a/3
<----------|-----------> f''(x)
- -a/3 +
So ranges above -a/3

B) The second derivative is a linear function so it only has one root.

C)so 0=-a/3 or a=0. To show that it has no critical points, plug a=0 into the first derivative.
f'(x)=3x^2+0+b
0=3x^2+b
3x^2=-b
This is impossible because it is given that b is positive and x^2 is positive.

Answer 2

f'(x) = 3x^2 + 2ax + b
f''(x) = 6x + 2a

A) f is concave up where f''(x) > 0.

6x + 2a > 0
6x > -2a
x > -1/3a

So f is concave up where x > -1/3 a

B) f has exactly one inflection point where f''(x) = 0. That point is x = -1/3a.

C) (0, -2) being the inflection point means that x = 0 is where the inflection point occurs. Therefore a = 0.

So:

f(x) = x^3 + bx + c

and

f(0) = 0^3 + b(0) + c = -2, so c = -2.

So now we have

f(x) = x^3 + bx - 2

and

f'(x) = 3x^2 + b

The critical point would be where f'(x) = 0

3x^2 + b = 0
3x^2 = -b

But you're given that b > 0, so you can't take the square root of it. Therefore, f has no critical points.