Question
Verify Rolle's theorem for the following function on the indicated intervals
f(x) = (x - 1) (x - 2)2 on [1, 2]
f(x) = (x - 1) (x - 2)2 on [1, 2]
f(x) = (x - 1)(x - 2)2
i.e. f(x) = x3 + 4x - 4x2 - x2 - 4 + 4x
i.e. f(x) = x3 - 5x2 + 8x - 4
We know that a polynominal function is everywhere derivable and hence continuous.
So, being a polynomial function, f(x) is continuous and derivable on [1, 2]
Also,
f(1) = (1)3 - 5(1)2 + 8(1) - 4 = 0
f(2) = (2)3 - 5(2)2 + 8(2) - 4 = 0
$\therefore$ f(1) = f(2) = 0
Thus, all the continuous of Rolle's theorem are satisfied.
Now, we have to show that there exists
$\text{c}\in(1,2)$ such that f'(c) = 0.We have,
f(x) = x3 + - 5x2 - 8x - 4
⇒ f'(x) = 3x2 + 8 - 10x
$\therefore$ f'(x) = 0 ⇒ 3x2 - 10x + 8 = 0
⇒ 3x2 - 6x - 4x + 8 = 0
⇒ 3x(x - 2) - 4(x - 2) = 0
⇒ (x - 2)(3x - 4)
$\Rightarrow\text{x}=2,\frac{4}{3}$
Thus, $\text{c}\in(1,2)$ such that f'(c) = 0.
Hence, Rolle's theorem is verified.
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