Question

Let f(x) = 5 + 12x − x^3. Find (a) the x- coordinate of all inflection points, (b)

the open intervals on which f is concave up, (c) the open intervals on which
f is concave down.

Answer 1

A) x = 0.

B) f is concave up for (-∞, 0).

C) f is concave down for (0, ∞).

Explanation:

We are given the function:

`f(x)=5+12x-x^3`

A)

We want to find the x-coordinates of all inflection points.

Recall that inflections points (may) occur when the second derivative equals zero. Hence, find the second derivative. The first derivative is given by:

`f'(x) = 12-3x^2`

And the second:

`f''(x) = -6x`

Set the second derivative equal to zero:

`0=-6x`

And solve for x. Hence:

`x=0`

We must test the solution. In order for it to be an inflection point, the second derivative must change signs before and after. Testing x = -1:

`f''(-1) = 6>0`

And testing x = 1:

`f''(1) = -6<0`

Since the signs change for x = 0, x = 0 is indeed an inflection point.

B)

Recall that f is concave up when f''(x) is positive, and f is concave down when f''(x) is negative.

From the testing in Part A, we know that f''(x) is positive for all values less than zero. Hence, f is concave up for all values less than zero. Our interval is:

`(-\infty, 0)`

C)

From Part A, we know that f''(x) is negative for all values greater than zero. So, f is concave down for that interval:

`(0, \infty)`