Question

Find ”c” of Lagrange’s mean value theorem for the function

f(x) = √(25-x^2) x ∈ [1,5]

Answer 1

Answer:

Answer

It is clear that f(x) has a definite and unique value of each x∈[1,5].

Thus, for every point in the interval [1, 5], the value of f(x) exists.

So, f(x) is continuous in the interval [1, 5].

Also, f

′

(x)=

25−x

2

−x

, which clearly exists for all x in an open interval (1, 5).

So, f

′

(x) is differentiable in (1,5).

So, there must be a value c∈[1,5] such that

f

′

(c)=

5−1

f(5)−f(1)

=

4

0−

24

=

4

0−2

6

=

2

−

6

But f

′

(c)=

25−c

2

−c

⇒

25−c

2

−c

=−

2

6

⇒4c

2

=6(25−c

2

)

⇒4c

2

=150−6c

2

⇒10c

2

=150

⇒c

2

=15⇒c=±

15

∴c=

15

∈[1,5]

Hence, option C is correct.