Suppose that 2 lessthanorequalto f\'(x) lessthanorequalto 4 for all values of x.

Question

Suppose that 2 lessthanorequalto f'(x) lessthanorequalto 4 for all values of x. What are the minimum and maximum possible values of f(5) - f(3)? lessthanorequalto f(5) - f(3) lessthanorequalto Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval f(x) 4x^2 + 4x + 4, [-1, 1] Yes, it does not matter if f is continuous or differentiable; every function satisfies the Mean Value Theorem. No, f is continuous on [-1, 1] but not differentiable on (-1, 1). No, f is not continuous on [-1 1]. There is not enough information to verify if this function satisfies the Mean Values Theorem. Yes, f is continuous on [-1, 1] and differentiable on (-1, 1) since polynomials are continuous and differentiable on R. If it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean Value Theorem (Enter your answers as a comma-separated list. If it does not satisfy the hypotheses, enter DNE.) c =

Explanation / Answer

4)given 2<= f '(x) <=4

so slope between any 2 points lies in interval [2,4]

slope betwee (3,f(3)),(5,f(5))  lies in interval [2,4]

2<= [f(5)-f(3)]/(5-3) <=4

2<= [f(5)-f(3)]/2 <=4

2*2<= f(5)-f(3) <=4*2

4<= f(5)-f(3) <=8

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5)f(x)=4x2+4x+4 is continous on [-1,1]and differentiable on (-1,1)  since it is a polynomial

f(-1)=4-4+4=4, f(1)=4+4+4=12

f(x)=4x2+4x+4

f '(x)=8x+4

f '(c)=8c +4

f '(c)=[f(1)-f(-1)]/(1-(-1))

8c +4=[12-4]/(2)

8c +4=8/2

8c +4=4

8c=0

c=0

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