Loading [MathJax]/jax/output/HTML-CSS/jax.js

Verify Rolle’s theorem for the function f(x) = x2 – 5x + 6 on the interval [2, 3].

Solution :

Since a polynomial function is everywhere differentiable and so continuous also.

Therefore, f(x) is continuous on [2, 3] and differentiable on (2, 3).

Also, f(2) = 22 – 5 × 2 + 6 = 0 and f(3) = 32 – 5 × 3 + 6 = 0

f(2) = f(3)

Thus, all the conditions of rolle’s theorem are satisfied. Now, we have to show that there exists some c (2, 3) such that f'(c) = 0.

for this we proceed as follows,

We have,

f(x) = x2 – 5x + 6 f'(x) = 2x – 5

f'(x) = 0 2x – 5 = 0 x = 2.5

Thus, c = 2.5 (2, 3) such that f'(c) = 0. Hence, Rolle’s theorem is verified.


Similar Questions

Find the approximate value of f(3.02), where f(x) = 3x2+5x+3.

If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximating error in calculating its volume.

It is given that for the function f(x) = x36x2+ax+b on [1, 3], Rolles’s theorem holds with c = 2+13. Find the values of a and b, if f(1) = f(3) = 0.

Find the point on the curve y = cos x – 1, x [π2,3π2] at which tangent is parallel to the x-axis.

Leave a Comment

Your email address will not be published. Required fields are marked *