Consider the equation below. (If an answer does not exist, enter DNE.)
f(x) = x4 − 32x2 + 6
(a) Find the interval 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.
local minimum value

Incorrect: Your answer is incorrect.
local maximum value

Incorrect: Your answer is incorrect.

(c) Find the inflection points.
(x, y) =

(smaller x-value)
(x, y) =

(larger x-value)

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.)

a. f is increasing in the interval and decreasing in the intervals

b.  local maximum value of the function is 6, -250 and Local minimum value of the function is -250

c.  inflexion points are

f is concave up in intervals and concave down in interval

Step-by-step explanation:

Given:

To find: interval on which f is increasing and decreasing, local minimum and maximum values of f, inflection points and interval on which f is concave up and concave down

Solution:

A function f is increasing in the interval in which and decreasing in the interval in which

If a function is increasing before a point and decreasing after that point, the point is said to be a point of local maxima.

If a function is decreasing before a point and increasing after that point, the point is said to be a point of local minima.

An inflection point is a point on the graph of a function at which the concavity changes. Put second derivative equal to zero as check if concavity changes at the points obtained. Such points are said to be points of inflexion.

A function f is concave up in the interval in which   and concave down in the interval in which

a.

Observe the attached table.

So, f is increasing in the interval and decreasing in the intervals

b.

From the table,

a function f has a local maxima at and local minima at

So, local maximum value of the function is 6, -250

Local minimum value of the function is -250

c.

See the attached table

So, f is concave up in intervals

and concave down in interval

Also,

So, inflexion points are

-3.396 = -3.4

-0.708 = -0.7

0.304 = 0.3

0.955 = 1

4 puzzles in 3 hours

step-by-step explanation: