1. Explain why Rolle\'s Theorem does not apply to the function even though there
ID: 2848319 • Letter: 1
Question
1. Explain why Rolle's Theorem does not apply to the function even though there exist a and b such that f(a)=f(b)
f(x)= |1/x|, [-1,1]
2.Find the two x-intercepts of the function f and show that f '(x)=0 at some point between the two x-intercepts
f(x)=x^2 - x- 2
3.Determine if rolles theorem can be applied to f on the closed interval [a,b]. If rolles theorem can be applied, find all values of c in the open interval (a,b) such that f '(c)=0. If rolles theorem can't be applied,explain why not.
f(x)= cos(x) [0,2pi]
4.Determine whether the mean value theorem can be applied to f on the closed interval [a,b]. If the mean value theorem can be applied, find all values of c in the open interval (a,b) such that f '(c)= f(b)-f(a)/b-a. If the mean value theorem can't be applied, explain why not.
f(x)=x^2, [-2,1]
5.Determine the open intervals on which the graph is concave upward or downward.
y=-3x^5 +40x^3 +135x/ 270
6.Find the points of inflection and discuss the concavity of the graph of the function.
f(x)=2 csc 3x/2 , (0,2pi)
7.Find all relative extrema. Use the second derivative test where applicable.
f(x)= x^4 -4x^3 +2
Explanation / Answer
1. Explain why Rolle's Theorem does not apply to the function even though there exist a and b such that f(a)=f(b)
f(x)= |1/x|, [-1,1]
given function is discontinuous at x=0 in the given interval [-1,1]
2.Find the two x-intercepts of the function f and show that f '(x)=0 at some point between the two x-intercepts
f(x)=x^2 - x- 2
intercepts = f(x) =0
so (x+1)(x-2) = 0
so intercepts are -1,2 and interval is = [-1,2]
f'(x) = 2x-1 = 0
so x=0.5 lies between [-1,2] hence proved
3.Determine if rolles theorem can be applied to f on the closed interval [a,b]. If rolles theorem can be applied, find all values of c in the open interval (a,b) such that f '(c)=0. If rolles theorem can't be applied,explain why not.
f(x)= cos(x) [0,2pi]
rolles theorem can be applied bcoz given function is continuous in the given interval [0,2pi]
f'(x) = -sin x = 0
x=0,pi,2*pi
x= pi lies between (0,2pi)
4.Determine whether the mean value theorem can be applied to f on the closed interval [a,b]. If the mean value theorem can be applied, find all values of c in the open interval (a,b) such that f '(c)= f(b)-f(a)/b-a. If the mean value theorem can't be applied, explain why not.
f(x)=x^2, [-2,1]
mean value theorem can be applied in the given interval bcoz its continuous
f'(c) = f(1)-f(-2) / (1-(-2)) = 1-4/1+2 = -1
f'(x) = 2x
f'(c)=2c= -1
so c = -0.5 belongs to (-2,1)
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