Calc Help Please!?

Question

Let f(x)= x^3 + ax^2 + bx + c where a, b, and c are constants with a > or = 0 and b > 0

(a) Over what intervals is f concave up? Concave down?
(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

f(x) = x^3 + ax^2 + bx + c, a >= 0 and b > 0.

To determine concavity, we need to solve for the second derivative and then make it 0.

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

Now, we make f''(x) equal to 0.

0 = 6x + 2a

And then we solve for x.

-6x = 2a, therefore x = (-1/3)a = -a/3

Our critical number is -a/3. To determine where this is concave up and down, we need to test one value less than -a/3, and one value greater than -a/3. We want to test it in this function:

f''(x) = 6x + 2a

For a value greater than -a/3, choose x = -a/3 + 1.
Then f''(x) = 6(-a/3 + 1) + 2a = (-2a + 6) + 2a = 6, which is positive. Therefore, we know that the function is concave up on (-a/3, infinity)

For a value less than -a/3, choose x = -a/3 - 1.
Then f''(x) = 6(-a/3 - 1) + 2a = (-2a - 6) + 2a = -6, which is negative. Therefore, we know the function is concave down on (-infinity, -a/3).

f is concave up on (-a/3, infinity), and
f is concave down on (-infinity, a/3)

(b) To show that f has exactly one inflection point, we refer back to our second derivative.

f''(x) = 6x + 2a

This can only get us AT MOST one solution. Since we have intervals of concavity, there has to be one inflection point; it's at
f(-a/3) = (-a/3)^3 + a(-a/3)^2 + b(-a/3) + c
f(-a/3) = -a^3/27 + a(a^2/9) - [ab]/3 + c
f(-a/3) = -a^3/27 + a^3/9 - [ab]/3 + c

(c)

We're given that f(0) = -2, so

f(0) = c = -2

Since we just solved for our inflection point general formula, and we're given that (0,-2) is an inflection point, it follows that
f''(0) = 0
f''(0) is also equal to 6x + 2a; therefore,
f''(0) = 6(0) + 2a = 0, or 2a = 0, or a = 0.

Therefore, a = 0, c = -2, and
f(x) = x^3 + (0)x^2 + bx - 2, or
f(x) = x^3 + bx - 2

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

To solve for the critical points, we make f'(x) equal to 0.

0 = 3x^2 + b
-b = 3x^2
-b/3 = x^2, therefore
x = "plus or minus" sqrt (-b/3)

HOWEVER, it is given that b is STRICTLY GREATER than 0, so it follows that -b/3 is a negative number. We can't take the square root of -b/3 with the given condition that it is greater than 0. If it were greater than or equal to, we'd get a solution (since we can take the square root of 0). But unfortunately, it is strictly greater.

Answer 2

a) Take derivatives

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

f''(x) = 6x + 2a

Now find the roots

x = -2a/6 = -a/3 >=0

Find the sign from f''

When its positive, f is concave up

b) I only find a root, so theres only one

c) If the inflexion point is (0,-2), then x = 0, then a = 0

Find c using f(0) = 2. Its quite easy. c = 2

No idea what critical points are, sorry...If they are minima and maxima, then find f'

f'(x) = x^2 + 2, which has no real roots

Ana

Answer 3

A. Is just another way of asking, "Where is the 2'nd derivative positive or negative." (Negative is concave up)

B. Show that there is only one point where where the 2'nd derivative is zero.

C. Do the calculation(s) and show that the 1'st derivative is nowhere zero.


Doug