Plz solve part 4 of question 4

(i) The given function is fx=3+x-223.

Differentiating with respect to x, we get

f'x=23x-223-1f'x=23x-2-13f'x=23x-213

Clearly, we observe that for x=21, 3f'x does not exist.

Therefore, fx is not derivable on 1, 3.

Hence, Rolle's theorem is not applicable for the given function.

(ii) The given function is fx=x.
The domain of is given to be -1, 1.

Let c-1, 1 such that is not an integer.
Then,
limxcfx=fc

Thus, fx is continuous at x=c.

Now, let c=0.

Then,

limx0-fx=-10=f0

Thus, is discontinuous at = 0.

Therefore, fx is not continuous in -1, 1.

Rolle's theorem is not applicable for the given function.

(iii) The given function is fx=sin1x.
The domain of is given to be -1, 1.

It is known that limx0sin1x does not exist.

Thus, fx is discontinuous at x = 0 on -1, 1.

Hence, Rolle's theorem is not applicable for the given function.


(iv) The given function is fx=2x2-5x+3 on 1, 3.
The domain of is given to be 1, 3.
It is a polynomial function.
Thus, it is everywhere derivable and hence continuous.

But
f1=0 and f3=6f3f1

Hence, Rolle's theorem is not applicable for the given function.

(v) The given function is fx=x23 on -1, 1.
The domain of is given to be -1, 1.

Differentiating fx with respect to x, we get

f'x=23x-13
We observe that at x=0f'x is not defined.
Hence, Rolle's theorem is not applicable for the given function.

(vi) The given function is 
fx=-4x+5, 0x12x-3, 1<x2

At x = 0, we have

limx1-fx=limh0f1-h=limh0-41-h+5=1
And
limx1+fx=limh0f1+h=limh021+h-3=-1

 limx1-fxlimx1+fx

Thus, fx is discontinuous at x=1.
Hence, Rolle's theorem is not applicable for the given function.

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