2 years ago

For what values of x is f(x)>0?

Mathematics

1 answer:

Hitman42 [59]2 years ago

5 0

Answer:

for x € (-6,-1)U(4,6) f(x) >0

Step-by-step explanation:

hope it helps you

You might be interested in

A shirt is on sale for $30. It was originally $40 What is the percent of the decrease?

SpyIntel [72]

Answer:

25% percent

Step-by-step explanation:

4 0

3 years ago

Jennifer's grandmother buys carrots at a farm stands she and her three children equally share the carrots the total weight of th

juin [17]

Answer:

jennifer will get alot of carrots

Step-by-step explanation: good eyesight

7 0

3 years ago

A swimming pool in the shape of a rectangular prism is 10 feet long, 20 feet wide, and 5 feet deep. How much water could the swi

sasho [114]

Answer:

1,000 units^3 of water

Step-by-step explanation:

20•10•5 =1,000

8 0

3 years ago

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an

Lapatulllka [165]

Answer:

Step-by-step explanation:

Given that there is a function of x,

$f(x) = 2sin x + 2cos x,0\leq x\leq 2\pi$

Let us find first and second derivative for f(x)

$f'(x) = 2cosx -2sinx\\f"(x) = -2sinx-2cosx$

When f'(x) =0 we have tanx = 1 and hence

a) f'(x) >0 for I and III quadrant

Hence increasing in $(0, \pi/2) U(\pi,3\pi/2)\\$

and decreasing in $(\pi/2, \pi)U(3\pi/2,2\pi)$

$x=\frac{\pi}{4}, \frac{3\pi}{4}$

$f"(\pi/4)$

Hence f has a maxima at x = pi/4 and minima at x = 3pi/4

b) Maximum value = $2sin \pi/4+2cos \pi/4 =2\sqrt{2}$

Minimum value = $2sin 3\pi/4+2cos 3\pi/4 =-2\sqrt{2}$

f"(x) =0 gives tanx =-1

$x= 3\pi/4, 7\pi/4$

are points of inflection.

concave up in (3pi/4,7pi/4)

and concave down in (0,3pi/4)U(7pi/4,2pi)

3 0

3 years ago

HELP ME WITH 6 PLEASE!!!!!

valkas [14]

The awnsers is C hope this helps

6 0

3 years ago