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consider the function f(x)= (e^x)/(4+(e^x))A) Then f'(x) = B) The interval of increase...

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masterpiece56 | Student | (Level 1) Honors

Posted April 25, 2012 at 4:27 AM via web

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consider the function f(x)= (e^x)/(4+(e^x))

A) Then f'(x) =

B) The interval of increase for f(x)
C) The interval of decrease for f(x)

D) f(x)  has a local minimum at
E) f(x) has a local maximum at
F) Then =
G) The interval of upward concavity for f(x) is
H) The interval of downward concavity for f(x) is
I) f(x) has an inflection value, x = ?

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jihyunstellakim | Elementary School Teacher | (Level 1) eNoter

Posted April 25, 2012 at 4:49 AM (Answer #1)

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a) f'(x)=((e^x)(4+e^x)-(e^x)(e^x))/((4+e^x))^2

         =4(e^x)/(4+e^x)^2

 

b) For f(x) to be incresing, f'(x)>0

Since 4(e^x)/(4+e^x)^2 is positive on all interval, the interval of increase is [-∞,∞]

c) For f(x) to be decreasing, f'(x)<0

There is no such value of x.

d) Critical points of x when f'(x)=0

f'(x) has no value of x that equals 0

therefore, there is no critical point, maximum or minimum.

same applies for e.

g) f''(x)= (-4(e^x)((e^x)-4))/((e^x)+4)^3

f(x) is concave up when f''(x)>0

f''(x)>0 when x>2ln(2)

Interval of upward concavity (2ln(2),∞)

h) f(x) is concave down when f''(x)<0

f''(x)<0 when x<2ln(2)

Interval of upward concavity (-∞,2ln(2))

i) inflection value when f''(x)=0

f''(x)=0 when x=2ln(2)

 
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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted April 25, 2012 at 8:32 AM (Answer #2)

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A) You need to find derivative of function using quotient law such that:

`f'(x) = ((e^x)'*(4+(e^x)) - (e^x)*(4+(e^x))')/((4+(e^x))^2)`

`f'(x) = ((e^x)*(4+(e^x)) - (e^x)*(e^x))/((4+(e^x))^2)`

`f'(x) = (4e^x + e^(2x) - e^(2x))/((4+(e^x))^2)`

Reducing like terms yields:

`f'(x) = (4e^x)/((4+(e^x))^2)`

B) Notice that `4e^x gt 0`  and `((4+(e^x))^2) gt 0` , hence f'(x)>0 for `x in (-oo,oo), ` hence, the function increases over R set.

C) The function only increases over R set.

D) Since `f'(x)!=0`  for any `x in R`  set, hence the function has no local minimum.

E) Since `f'(x)!=0`  for any `x in R`  set, hence the function has no local maximum either.

G) You need to evaluate f''(x) to find where the function is concave up or concave down such that:

`f''(x) = ((4e^x)'*((4+(e^x))^2) - (4e^x)*((4+(e^x))^2)')/((4+(e^x))^4)`

`f''(x) = (4e^x*((4+(e^x))^2) - 2e^x(4e^x)*((4+(e^x)))/((4+(e^x))^4)`

You need to factor out `(4e^x)*((4+(e^x))`  such that:

`f''(x) = (4e^x)*((4+(e^x))(4 + e^x - 2e^x)/((4+(e^x))^4)`

`f''(x) = (4e^x)*(4- e^x)/((4+(e^x))^3)`

You need to solve f''(x) = 0 such that:

`(4e^x)*(4 - e^x)/((4+(e^x))^3) = 0 =gt (4e^x)*(4 - e^x) = 0`

Since `4e^x gt 0 =gt 4 - e^x = 0 =gt e^x = 4 =gt x = ln 4`

If `x lt ln 4 =gt f''(x) gt 0` , hence the function is concave up over `(-oo,ln 4).`

H) Notice that for `x gt ln 4 =gt f''(x) lt 0` , hence the function is concave down over `(ln 4, oo).`

I) Notice that the function has an inflection point at`x = ln 4` .

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