Mathematics, 29.05.2020 00:02 edamite12

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.)

Answers: 2

Show answers

Answers

a. f is increasing in the interval (-4,0) and decreasing in the intervals $\left ( -\infty ,-4 \right )\,,\,\left ( 0,4 \right )$

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

c. inflexion points are $\left ( \frac{-4\sqrt{3}}{3},\frac{-1226}{9} \right )\,,\,\left ( \frac{4\sqrt{3}}{3},\frac{-1226}{9} \right )$

f is concave up in intervals $\left ( -\infty ,\frac{-4\sqrt{3}}{3} \right )\,,\,\left ( \frac{4\sqrt{3}}{3},\infty \right )$ and concave down in interval $\left (\frac{-4\sqrt{3}}{3},\frac{4\sqrt{3}}{3} \right )$

Step-by-step explanation:

Given: f(x)=x^4-32x^2+6

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 f'0 and decreasing in the interval in which f''

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 f''0 and concave down in the interval in which f''

$f(x)=x^4-32x^2+6\\f'(x)=4x^3-64x\\=4x(x^2-16)\\=4x(x+4)(x-4)$

$f'(x)=0\\4x(x+4)(x-4)=0$

Observe the attached table.

So, f is increasing in the interval (-4,0) and decreasing in the intervals $\left ( -\infty ,-4 \right )\,,\,\left ( 0,4 \right )$

From the table,

a function f has a local maxima at x=0,4 and local minima at x=-4

$f(-4)=(-4)^4-32(-4)^2+6=256-512+6=-250\\f(0)=0^4-32(0)^2+6=6\\f(4)=4^4-32(4)^2+6=256-512+6=-250$

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

Local minimum value of the function is -250

$f'(x)=4x^3-64x\\f''(x)=12x^2-64=4(3x^2-16)\\f''(x)=0\Rightarrow 4(3x^2-16)=0\\3x^2-16=0\\x^2=\frac{16}{3}\\x=\pm \frac{4}{\sqrt{3}}=\pm \frac{4\sqrt{3}}{3}$

See the attached table

So, f is concave up in intervals $\left ( -\infty ,\frac{-4\sqrt{3}}{3} \right )\,,\,\left ( \frac{4\sqrt{3}}{3},\infty \right )$

and concave down in interval $\left (\frac{-4\sqrt{3}}{3},\frac{4\sqrt{3}}{3} \right )$

Also,

$f\left ( \frac{-4\sqrt{3}}{3} \right )=\left ( \frac{-4\sqrt{3}}{3} \right )^4-32\left ( \frac{-4\sqrt{3}}{3} \right )^2+6=\frac{-1226}{9}\\f\left ( \frac{4\sqrt{3}}{3} \right )=\left ( \frac{4\sqrt{3}}{3} \right )^4-32\left ( \frac{4\sqrt{3}}{3} \right )^2+6=\frac{-1226}{9}$

So, inflexion points are $\left ( \frac{-4\sqrt{3}}{3},\frac{-1226}{9} \right )\,,\,\left ( \frac{4\sqrt{3}}{3},\frac{-1226}{9} \right )$

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

Answer from: Quest

hey there!

if she's saving for six months, let's divide the total number we need for the trip by six, dividing it into the amount that needs to be saved for each month.

500 ÷ 6 = $83.\overline3$

she needs to save at least this amount each month, so we should use this symbol: ≥

let x be the amount she saves each month.

your final answer is x ≥ $83.\overline3$

hope this !

Answer from: Quest

u bout ugly as a mf

step-by-step explanation: lol

Which set of angles are vertical angles? a) 1 and 7 b) 1 and 4 c) 2 and 4 d) 3 and 5

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

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