How do you determine if rolles theorem can be applied to #sqrt(x) - 1/5x # on the interval [0,25] and if so how do you find all the values of c in the interval for which f'(c)=0?
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Rolle's Theorem has three hypotheses:
H1 : #f# is continuous on the closed interval #[a,b]#
H2 : #f# is differentiable on the open interval #(a,b)#.
H3 : #f(a)=f(b)#
In this question, #f(x) = sqrtx - 1/5x# , #a=0# and #b=25#.
We can apply Rolle's Theorem if all 3 hypotheses are true.
H1 : This function is continuous on its domain, #[0, oo)#, so it is continuous on #[0, 25]#
H2 : #f'(x)=1/(2sqrt(x-7))# which exists for all #x >0# (for all #x in (0,oo)#,
So #f# is differentiable on #(0, 25)#
H3 : #f(0) = 0 = f(25)#
Therefore we can apply Rolle's Theorem to #f(x) = sqrtx - 1/5x# on the interval: #[0, 25]#
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As an algebra exercise you've also been asked to actually find the values of #c# whose existence is guaranteed by Rolle's Theorem.
To do that, find #f'(x)#, set it equal to #0# and do the algebra to solve the equation. Select any solutions in #(0, 25)#.
