If f(a)=f(b)=0 and f(x) is continous on [a,b], then

a. f(x) must be identically 0
b. f' (x) may be different from zero for all x on [a,b]
c. there exists at least one number c, a d. f' (x) must exist for every x on (a,b)
e. none of the preceding is true.

Calculus Help.

say f(x)- a circle x^2+y^2=4 and [a,b]=[-2,2]( assumption is circle and the diametric ends)

a. no need for f[x]=0 all the time
b. f`(X)=Ax (A != 0) which is equal to 0 when x=0
c. yes when we see the above bit x=0 f`(x)=0 -2<0<2 (ANSWER IS C)

The answer you chose is wrong! That person used an example that worked, but what about f(x)=-|x|+4? (c) is not right at all. There is a difference between differentiability and continuity. If you assume differentiability, then (c) and (d) are both right. I got a 5 on AP Calculus as a sophomore.

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I agree with the first commentor, C) is not the answer. The answer is B) and the reason is the same function f(x) = - |x|+4. You have f(-4) = f(4) = 0, f(x) is continuous but f'(x) is never zero in between -4 and +4. There is a turning point at zero, but f(x) is not differentiable due to the corner!

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(a) is not true, because there are functions that cross the x-axis twice (f(x)=x^2-1)
(b) is true, because there could be a cusp
(c) is not true, because there could be a cusp
(d) is not true, because there could be a cusp, where f is continuous but not differentiable

Rolle's theorem does not apply because the equation is not specified as being differentiable, only continuous.

See Rolle´s Theorem .- Answer C and D.-
C by Rolle´s Theorem , D because the curve is continous as they say and because it is one condition of Rolle´s Theo)