Question

Verify Mean Value Theorem, if f(x)=x^3-5x^2-3x in the interval [a, b], where a = 1 and b = 3. Find all c∈(1,3) for which f'(c)=0

Answer 1

Mean Value Theorem states that for a function f : [a, b] → R, if
(a)f is continuous on [a, b]
(b)f is differentiable on (a, b)
Then there exists some c ∈ (a, b) such that
$\bf{f'(c)=\frac{f(b)-f(a)}{b-a}}$

Given, f(x) = x³ - 5x² - 3x
As f(x) is a polynomial function,
(a) f(x) is continuous in [1, 3]
(b) f'(x) = 3x² - 10x - 3
So, f(x) is differentiable in (1, 3).
so, f'(c) = 3c² - 10c - 3 = {f(3) - f(1)}/(3 - 1)
f(3) = 3³ - 5(3)² - 3(3) = 27 - 45 - 9 = -27
f(1) = 1³ - 5(1)² - 3.1 = 1 - 5 - 3 = -7
now, 3c² - 10c - 3 = (-27 + 7)/2
=> 3c² - 10c - 3 = -10
=> 3c² - 10c + 7 = 0
=> 3c² - 3c - 7c + 7 = 0
=> 3c(c - 1) - 7(c - 1) = 0
=> (3c - 7)(c - 1) = 0
=> c = 7/3 , 1
where c = 7/3 ∈ (1, 3)
The Mean Value Theorem is verified for the given f(x) and c = 7/3 ∈ (1, 3) is the only point for which f'(c) = 0.

Answer 2

Answer:

The Mean Value Theorem is verified for the given f(x) and c = 7/3 ∈ (1, 3) is the only point for which f'(c) = 0.