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
![f^{l} (c) = \frac{f(b) -f(a)}{b-a}](https://tex.z-dn.net/?f=f%5E%7Bl%7D%20%28c%29%20%20%3D%20%5Cfrac%7Bf%28b%29%20-f%28a%29%7D%7Bb-a%7D)
<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
![f^{l} (c) = \frac{f(0) -f(-1)}{0-(-1)}](https://tex.z-dn.net/?f=f%5E%7Bl%7D%20%28c%29%20%20%3D%20%5Cfrac%7Bf%280%29%20-f%28-1%29%7D%7B0-%28-1%29%7D)
![8c+4 = \frac{-3-(4(-1)^2+4(-1)-3)}{0-(-1)}](https://tex.z-dn.net/?f=8c%2B4%20%3D%20%5Cfrac%7B-3-%284%28-1%29%5E2%2B4%28-1%29-3%29%7D%7B0-%28-1%29%7D)
8c+4 = -3-(-3)
8c+4 = 0
8c = -4
![c = \frac{-4}{8} = \frac{-1}{2} = -0.5](https://tex.z-dn.net/?f=c%20%3D%20%5Cfrac%7B-4%7D%7B8%7D%20%3D%20%5Cfrac%7B-1%7D%7B2%7D%20%3D%20-0.5)
c ∈ (-1,0)
<u>Conclusion</u>:-
The value of c = -0.5∈ (-1,0)
<u></u>
Answer:
27
Step-by-step explanation:
3*3*3=27
Answer:
-1
Step-by-step explanation:
Assuming problem is: x->2 (x^3 -3x^2 +3)
Since the expression is continuous at x=2,
then the limit can be found just by evaluating the expression at x=2.
2^3-3(2)^2+3
8-3(4)+3
8-12+3
-4+3
-1