Answer:
2 i think
Step-by-step explanation:
Answer:
Domain = -2, range = -5
Domain =>0, range = 0
domain = 2, range = 5
let y =>f(x)
It can be seen that Function can be linear having common difference 2
F(x) = 2x + 1 for x>0
is satisfied for above values
let's check
F(2) = 2(2)+1 = 4+1 = 5 As given
For x<0 , F(x) = 2x -1
f(-2)= 2(-2) -1= -4-1= -5 As given
For X= 0, F(x)= 2x
So , F(0) = 0 As given
The required expression is 
Step-by-step explanation:
We need to identify the expression correctly represents “nine less than the quotient of a number and four, increased by three"
Solving:
Let the number is x
The quotient of number and four: 
nine less than the quotient of a number and four: 
Increased by 3 : 
So, nine less than the quotient of a number and four, increased by three is:

So, The required expression is 
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Answer:
Step-by-step explanation:
Given a function f, whose derivatives are f' and f'', a value x is a critical point if f'(x) =0. A value x is a minimum of f if it is a critical point and f''(x) >0 and it is maximum if f''(x)<0. We will perfom the following steps:
1. Calculate the derivative f'.
2. Solve f'(x) =0.
3. Determine if the x value found in 2 is a minimum or a maximum using f''.
Recall the following properties of derivatives


where c is a constant.

where f,g are differentiable.
where c is a constant.
(chain rule)
Case 1: f(x) = 2+3x+3.
Using the properties from above, we have
1.
2. The equation f'(x)=0 where f'(x) = 3 has no solution.
3. Based on the previous result, f has no maximum nor minimum.
Case 2: 
1. 
2. We have the equation

which is equivalent to

Recall that the cosine function only takes values in the set [-1,1]. So, this equation has no solution.
3. Based on the previous result, f has no maximum nor minimum.