In this page Lagrange theorem questions solution4 we are going to see solution of the practice questions.
(2) If f (1) = 10 and f ' (X) ≥ 2 for 1 ≤ x ≤ 4 how small can f (4) possibly be?
Solution:
If f (x) be a real valued function that satisfies the following conditions.
1. |
f(x) is defined and continuous on the closed interval [1,4] |
2. |
f(x) is differentiable on the open interval (1,4). |
Then there exists at least one point c ∊ (1,4) such that f ' (c) = f (b) - f (a) / (b - a) |
f ' (c) = f (b) - f (a) / (b - a)
f ' (c) = [f (4) - f (1)] / (b - a)
f ' (c) = [f (4) - 10] / (4 - 1)
f ' (c) = [f (4) - 10]/3
3 f ' (c) = [f (4) - 10]
3 f ' (c) + 10 = f (4)
f (4) ≥ 6 + 10
f (4) ≥ 16. So the minimum value of f (4) must be 16.
(3) At 2.00 p.m car's speedometer reads 30 miles/hr., at 2. 10 pm it reads 50 miles/hr. Show that sometime between 2.00 and 2.10 the acceleration is exactly 120 miles/hr²
Solution:
Let velocity be "v" at time "t"
1. |
f(x) is defined and continuous on the closed interval [2,2.10] |
2. |
f(x) is differentiable on the open interval (2,2.10). |
v (2) = 30 miles/hour v (2.10) = 50 miles/hour we can find the value of c by using the above condition v '(c) = [v (2.10) - v (2)]/2.10 - 2 = [50 - 30]/(10/60) miles/hour = 20/(10/60) miles/hour = 20 x (60/10) = 120 miles/hour |