Question

The value of c in the mean value theorem f(b) – f(a) = (b – a) f ′(c)

where f(x) = Ax2 + Bx + C in (a, b) is ​

Answer 1

Answer:

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

3