Answer:
The value of c = -0.5∈ (-1,0)
Step-by-step explanation:
<u>Step(i)</u>:-
Given function f(x) = 4x² +4x -3 on the interval [-1 ,0]
<u> Mean Value theorem</u>
Let 'f' be continuous on [a ,b] and differentiable on (a ,b). The there exists a Point 'c' in (a ,b) such that

<u>Step(ii):</u>-
Given f(x) = 4x² +4x -3 …(i)
Differentiating equation (i) with respective to 'x'
f¹(x) = 4(2x) +4(1) = 8x+4
<u>Step(iii)</u>:-
By using mean value theorem


8c+4 = -3-(-3)
8c+4 = 0
8c = -4

c ∈ (-1,0)
<u>Conclusion</u>:-
The value of c = -0.5∈ (-1,0)
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