Give me the answer in francais
Answers
Explanation:
Value of θ mean value theorem
f(x)=ax
2
+bx+c in [1,2]
→f(x)=ax
2
+bx+c in [a,b]
Where a=7,b=2
→ Condition of mean value theorem
(i) f(x) is continuous at (a,b)
(ii) f(x) is derivable at (a,b)
If both condition satisfied then there exist some θ vallue in (a,b) such that f
′
(θ)=
b−a
f(b)−(a)
→ condition : 1
f(x)=ax
2
+bx+c
f(x) is polynomial & every polynomial f
n
is continuous
f(x) is continuous at x∈(1,7)
→ Condition 2
f(x)=ax
2
+bx+c
f(x) is polynomial & every polynomial f
n
is differentiable
-so, f(x) is differentiable at x∈(1,2)
Now f(x)=ax
2
+bx+c
f
′
(x)=2ax+b
f
′
(θ)=2aθ+b
Also, f(1)=a(1)
2
+b(1)+c=a+b+c
f(2)=a(2)
2
+b(a)+c=4a+2b+c
By theorem f
′
(θ)=
b−a
f(b)−f(a)
=
2−1
f(2)−f(1)
∴24θ+b=4a+2b+c−(a+b+c)
24θ+b=3a+b
∴24θ=3a
θ=
2
hope it helps you
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