Consider the equation below. f ( x ) = 4 x 3 + 21 x 2 294 x + 8 (a) Find the int
ID: 2849165 • Letter: C
Question
Consider the equation below.
f(x) = 4x3 + 21x2 294x + 8
(a) Find the intervals on which f is increasing. (Enter your answer using interval notation.)
Find the interval on which f is decreasing. (Enter your answer using interval notation.)
(b) Find the local minimum and maximum values of f.
(c) Find the inflection point.
(x, y) =
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.)
Explanation / Answer
First thing, is increasing/decreasing intervals. These are marked by the critical points, which are where the derivative of f(x), aka f'(x) = 0.
Let's first find the derivative. f(x) = 4x^3 + 21x^2 - 294x + 7
f'(x) = 12x^2 + 42x -294 This must be set = 0 to find the x's
where f'(x) = 0. 12x^2 + 42x - 294 = 0
Now solve for 0. 6(2x^2 + 7x - 99) = 0 (divide both sides by the 6 we pulled out)
2x^2 + 7x - 99 = 0
Using the quadratic formula, we find that x = 5.5 and x = -9.
Refer to an algebra text for how I did this if you are not familiar with this formula. These points are where the increasing/decreasing changes (also known as critical points. We must test points that are outside and between these points to find behavior.
Let us test -10 for the (-infinity, -9) interval, 0 for the (-9,5.5) interval, and 6 for the (5.5, infinity) interval. Plug these values into f'(x) and look for the sign (+ or -) of the answer.
2(-10)^2 + 7(-10) - 99 200 - 70 - 99 = 69 which is POSITIVE, and so the interval (-infinity, -9) is INCREASING 2(0)^2 + 7(0) - 99 0 + 0 -99 = -99 which is NEGATIVE,
and so the interval (-9, 5.5) is DECREASING 2(6)^2 + 7(6) - 99 72 + 42 - 99 = 35 which is POSITIVE,
and so the interval (5.5, infinity) is INCREASING.
b). We now know there is a local MIN at 5.5 and a local MAX at -9. Next, we have to find concavity, based on the inflection points, which are done in the same fashion but using f"(x), the second derivative.
First we find the second derivative: f'(x) = 2x^2 + 7x - 99 f"(x) = 4x + 7.
Again, solve for x when f"(x) = 0. 4x + 7 = 0 4x = -7 x = -7/4.
c). This is our only INFLECTION POINT. We chose a value on each side, such as -2 for the interval (-infinity, -7/4) and 0 for the interval (-7/4,infinity).
Plug these into the f"(x) equation: 4x + 7 and get 4(-2) + 7 = -1 which is NEGATIVE so the interval (-infinity, -7/4) is concave DOWN.
and 4(0) + 7 = 7 which is POSITIVE so the interval (-7/4, infinity) is concave UP.
This is the same process for all questions asking for inc/dec, max/min, inflection pt, and concavity.