Answer:
A) x = 0.
B) f is concave up for (-∞, 0).
C) f is concave down for (0, ∞).
Step-by-step explanation:
We are given the function:
[tex]f(x)=5+12x-x^3[/tex]
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:
[tex]f'(x) = 12-3x^2[/tex]
And the second:
[tex]f''(x) = -6x[/tex]
Set the second derivative equal to zero:
[tex]0=-6x[/tex]
And solve for x. Hence:
[tex]x=0[/tex]
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:
[tex]f''(-1) = 6>0[/tex]
And testing x = 1:
[tex]f''(1) = -6<0[/tex]
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:
[tex](-\infty, 0)[/tex]
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:
[tex](0, \infty)[/tex]
