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Asked by Abdul Last Modified
Answer 
Sadika
To verify Rolle's Theorem for the function y=x2+2y=x2+2 on the interval [−2,2][−2,2], we need to check three conditions:
Let's check each condition:
Continuity: The function y=x2+2y=x2+2 is a polynomial function and is continuous everywhere. Therefore, it is continuous on the closed interval [−2,2][−2,2].
Differentiability: The function y=x2+2y=x2+2 is a polynomial function and is differentiable everywhere. Therefore, it is differentiable on the open interval (−2,2)(−2,2).
Endpoints: y(−2)=(−2)2+2=4+2=6y(−2)=(−2)2+2=4+2=6 y(2)=(2)2+2=4+2=6y(2)=(2)2+2=4+2=6
So, y(−2)=y(2)y(−2)=y(2).
Since all three conditions of Rolle's Theorem are satisfied, there exists at least one cc in the open interval (−2,2)(−2,2) such that f′(c)=0f′(c)=0, where f(x)=x2+2f(x)=x2+2.
Now, let's find the derivative of y=x2+2y=x2+2:
y′=ddx(x2+2)=2xy′=dxd(x2+2)=2x
To find the critical point(s), we set y′=0y′=0: 2x=02x=0 x=0x=0
So, f′(0)=0f′(0)=0.
Thus, Rolle's Theorem is verified for the function y=x2+2y=x2+2 on the interval [−2,2][−2,2], and there exists at least one point cc in the open interval (−2,2)(−2,2) where the derivative f′(c)f′(c) is equal to zero.
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