Consider the function below. g(x) = 190+8x^3+x^4 (a) Find the interval of increa
ID: 3192807 • Letter: C
Question
Consider the function below. g(x) = 190+8x^3+x^4 (a) Find the interval of increase. (Enter your answer in interval notation.) Find the interval of decrease. (Enter your answer in interval notation.) (b) Find the local minimum value(s). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) Find the local maximum value(s). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) (c) Find the interval where the graph is concave upward. (Enter your answer in interval notation.) Find the interval where the graph is concave downward. (Enter your answer in interval notation.) Find the inflection points. (x,y) = ( ) (smaller x-value) (x,y) = ( ) (larger x-value)Explanation / Answer
a)
to find the increase interval, we need to find the critical point by finding first derivative and equal it to zero as the following:
g '(x) = 24x^2 + 4x^3
0 = 24x^2 + 4x^3
0 = 4x^2 * ( 6 + x ) ----> x = 0 & -6
Now, we are going to check before and after the critical points into the first derived equation:
g '(x) = 24x^2 + 4x^3
---------+++++++++++++
_________________
. . . .-6. . . . . 0. . .
a) increase [-6 , )
decrease (- , -6]
b)
Now, we are going to plug the critical point into the original equation:
g(-6) = 190 + 8 * (-6)^3 + (-6)^4 = -242 local minimum critical value
g(0) = 190 + 8 * (0)^3 + (0)^4 = 190 Niether critical value
c)
Now, second derivative for points of inflection:
g ''(x) = 48x + 12x^2
0 = 12x * ( 4 + x ) ----> x = 0 & -4
Checking second derived equation:
g ''(x) = 48x + 12x^2
++++++--------++++++
_________________
. . . . -4 . . . . 0. . .
concave up (- , -4] U [0 , )
concave down [ -4 , 0 ]
inflection points by plugging into the original equation:
g(-4) = 190 + 8 * (-4)^3 + (-4)^4 = -66 local minimum
g(0) = 190 + 8 * (0)^3 + (0)^4 = 190 inflection point
(x,y) = (-4 , -66)
(x,y) = (0 , 190)
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