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Part(i):
The function f(x) is continuous in a closed interval (2<=x<=5).
Part(ii)
The function f(x) is differentiable in an open interval (2Part(iii)
Verify if f(2)=f(5)
f(x)=x²-7x+10
f(2)=(2-2) (2-5)=0
f(5)=(5-2) (5-5)=0
f(2)=f(5)=0 and therefore there exists a point C in the (a f(x)=x²-7x+10
f'(x)=2x-7
f'(c)=2c-7
0=2c-7
c=7/2
(2<=c<=5)
Kibet Koina answered the question on January 23, 2019 at 12:16
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