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wdymRachel411
29.05.2020 •
Mathematics
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.)
Solved
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Ответ:
a. f is increasing in the interval
and decreasing in the intervals ![\left ( -\infty ,-4 \right )\,,\,\left ( 0,4 \right )](/tpl/images/0669/4769/9d13d.png)
b. local maximum value of the function is 6, -250 and Local minimum value of the function is -250
c. inflexion points are![\left ( \frac{-4\sqrt{3}}{3},\frac{-1226}{9} \right )\,,\,\left ( \frac{4\sqrt{3}}{3},\frac{-1226}{9} \right )](/tpl/images/0669/4769/7d470.png)
f is concave up in intervals
and concave down in interval ![\left (\frac{-4\sqrt{3}}{3},\frac{4\sqrt{3}}{3} \right )](/tpl/images/0669/4769/7f3b0.png)
Step-by-step explanation:
Given:![f(x)=x^4-32x^2+6](/tpl/images/0669/4769/3eee1.png)
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 ![f''](/tpl/images/0669/4769/54539.png)
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 ![f''](/tpl/images/0669/4769/54539.png)
a.
Observe the attached table.
So, f is increasing in the interval
and decreasing in the intervals ![\left ( -\infty ,-4 \right )\,,\,\left ( 0,4 \right )](/tpl/images/0669/4769/9d13d.png)
b.
From the table,
a function f has a local maxima at
and local minima at ![x=-4](/tpl/images/0669/4769/14c42.png)
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![\left ( -\infty ,\frac{-4\sqrt{3}}{3} \right )\,,\,\left ( \frac{4\sqrt{3}}{3},\infty \right )](/tpl/images/0669/4769/d71fd.png)
and concave down in interval![\left (\frac{-4\sqrt{3}}{3},\frac{4\sqrt{3}}{3} \right )](/tpl/images/0669/4769/7f3b0.png)
Also,
So, inflexion points are![\left ( \frac{-4\sqrt{3}}{3},\frac{-1226}{9} \right )\,,\,\left ( \frac{4\sqrt{3}}{3},\frac{-1226}{9} \right )](/tpl/images/0669/4769/7d470.png)
Ответ:
The answer is c)5