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