Consider the equation below. f(x) = 4x^3 + 18x^2 − 216x + 1 (a) find the intervals on which f is increasing. find the interval on which f is decreasing. (b) find the local minimum and maximum values of f. (c) find the inflection point. find the interval on which f is concave up. (enter your answer using interval notation.) find the interval on which f is concave down. (enter your answer using interval notation.)
(-5, 0) ∪ (5, ∞)
I find a graph convenient for this purpose. (See below)
When you want to find where a function is increasing or decreasing, you want to look at the sign of the derivative. Here, the derivative is ...
f'(x) = 4x^3 -100x = 4x(x^2 -25) = 4x(x +5)(x -5)
This has zeros at x=-5, x=0, and x=5. The sign of the derivative will be positive when 0 or 2 factors have negative signs. The signs change at the zeros. So, the intervals of f' having a positive sign are (-5, 0) and (5, ∞).
(a) The function is increasing and decreasing
(b) The local minimum is x = 5 and the maximum is x = -6
(c) The inflection point is
(d) The function is concave upward on and concave downward on
(a) To find the intervals where is increasing or decreasing you must:
1. Differentiate the function
2. Now we want to find the intervals where f'(x) is positive or negative. This is done using critical points, which are the points where f'(x) is either 0 or undefined.
These points divide the number line into three intervals:
, , and
Evaluate f'(x) at each interval to see if it's positive or negative on that interval.
Therefore f(x) is increasing and decreasing
(b) Now that we know the intervals where f(x) increases or decreases, we can find its extremum points. An extremum point would be a point where f(x) is defined and f'(x) changes signs.
We know that:f(x) increases before x = -6, decreases after it, and is defined at x = -6. So f(x) has a relative maximum point at x = -6.f(x) decreases before x = 5, increases after it, and is defined at x = 5. So f(x) has a relative minimum point at x = 5.
(c)-(d) An Inflection Point is where a curve changes from Concave upward to Concave downward (or vice versa).
Concave upward is when the slope increases and concave downward is when the slope decreases.
To find the inflection points of f(x), we need to use the f''(x)
We set f''(x) = 0
Analyzing concavity, we get
The function is concave upward on because the f''(x) > 0 and concave downward on because the f''(x) < 0.
f(x) is concave down before , concave up after it. So f(x) has an inflection point at .
the answer is (-4 , 0) ∪ (4 , infinity)
f is decreasing if f '(x) = 4x^3 - 64x <0, so it is the same of x(x^2 -16)<0, implies x<0 or (x-4)(x+4)<0, so x< + or -4
the answer is (- infinity, -4) ∪ (0 , 4)
the local minimum and maximum
f '(x) =0, impolies x=+ or -4, or x=0
or f'(o)=0, and f'(- 4)=f'(4)= 0,
M(-4, 0) or M(4, 0) or M(0,0)
inflection points can be found by solving f '' (x)=12x^2 - 64 =0
x=+ or - 4sqrt(3) / 3
so the inflection point is and M(- 4sqrt(3) / 3, f'' ( -4sqrt(3)) (smaller x value), and M(4sqrt(3) / 3, f'' (4sqrt(3)) (larger x value)
f is concave up if f ''>0
it means 12x^2 - 64>0, so the interval is (- infinity, -4sqrt(3) U (4sqrt(3), infinity)
f is concave down if f ''<0
it means 12x^2 - 64<0 so the interval is (-4sqrt(3)) U (4sqrt(3))