Let A(x) = x squareroot/x + 5. Answer the following questions. Find the interval

Question

Let A(x) = x squareroot/x + 5. Answer the following questions. Find the interval(s) on which A is increasing. Answer (in interval notation): Find the interval(s) on which A is decreasing. Answer (in interval notation): Find the local maxima of A. List your answers as points in the form (a, b). Answer (separate by commas): Find the local minima of A. List your answers as points in the form (a, 6). Answer (separate by commas): Find the interval(s) on which A is concave upward. Answer (in interval notation): Find the interval(s) on which A is concave downward. Answer (in interval notation):

Explanation / Answer

A(x) = x*sqrt(x+5)
Difefrentiate using product rule,
A'(x) = 1*sqrt(x+5) + x/(2*sqrt(x+5))
            =(2(x+5)+x) / (2*sqrt(x+5))
            = (3x+10) / (2*sqrt(x+5))

put A'(x) = 0
3x+10 = 0
x = -10/3

for x < -10/3, A'(x) is negative
for x > -10/3, A'(x) is positive
so,
x = -10/3 is minima
There is no maxima

Minimum value = A(x) = (-10/3) * sqrt(-10/3 + 5)
    = (-10/3) * sqrt(5/3)
    = -4.3

Answer:

1.
Increasing in ,
x = (-10/3 , infinity)

2.
x+5>= 0
x>=-5
Since square root of negative number is not defined

decreasing in:
x = (-5, -10/3)

3.
local maxima at end point which is x=-5
A(x) = -5*sqrt(-5+5) = 0
Answer: (-5,0)

4.
local minima at:
(-10/3,-4.3)

only 4 parts at a time please

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