1) Determine the intervals where f(x) = 5/7x^7/5 is concave up and concave down
ID: 3343154 • Letter: 1
Question
1) Determine the intervals where f(x) = 5/7x^7/5 is concave up and concave down
Determine any points of inflection for f
2) Find the critical points for f(x) = x +cos(x), between the invervals 0 < or = x and < or = to 2pi
Determine the intervals where f is increasing or decreasing
Classify each critical point as local maximum, local minimum, or neither one
Determine the intervals where f is concave up and where it is concave down
Determine any points of in%uFB02ection for f
Please answer neatly...Thank You!!!
Explanation / Answer
f(x) = (5/7)x^(7/5)
Domain of f(x) is for x>0
f '(x) = (5/7)(7/5) x^(2/5) = x^(2/5)
1.)
For concave up f''(x)>0
f ''(x) = (2/5)(x^(-3/5)) = (2/5)/(x^(3/5))which is always greater than 0 for x>0
Concave up interval = [0,infinity)
Concave down = nullset
Point of inflection for f is value of x where f ''(x) =0
f '(x)= x^(2/5)=0 at x=0
But f ''(x) at x=0, is infinity
Thus no point of inflection exists
2.)
a.)
f(x) = x +cos(x)
f '(x) = 1 - sin(x) = 0 or sin(x) = 1
x=pie/2,
b.)
f '(x) = 1 - sin(x)
-1<sin(x) <1 for all 0<=x<=2pie
Thus 1-sinx >0 ,for all 0<=x<=2pie
Thus f is increasing for x = [0,2pie]
decreasing= nullset
c.)
There is only 1 critical point at x=pie/2,
f "(x) = d(1 - sin(x))/dx = -cos(x) = 0
Thus the critical point x=pie/2 is neither maxima nor minima.. It is a point of inflection
d.)
For concave up
f "(x) = -cos(x) >0 or cos(x) <0
x belongs to the interval (0,pie/2) union (3pie/2,2pie)
For concave down
f "(x) = -cos(x) <0 or cos(x) >0
x belongs to the interval (pie/2,3pie/2)
e.) Point of inflection for f is value of x where f ''(x) =0
f "(x) = -cos(x)=0 for x=pie/2 and 3pie/2
Thus the point of infection are x=pie/2 , 3pie/2
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