Question
Verify Rolle's theorem for the following function on the indicated intervals
f(x) = (x2 - 1)(x - 2) on [-1, 2]
f(x) = (x2 - 1)(x - 2) on [-1, 2]
f(x) = (x2 - 1)(x - 2) on [-1, 2]
f(x) is continuous is [-1, 2] and differentiable in (-1, 2) as it is a polynomial functions.
Now,
f(-1) = (1-1)(-1-2) = 0
f(2) = (4-1)(2-2) = 0
⇒ f(-1) = f(2)
So, Rolle's theorem is applicable on f(x) is [-1, 2] therefore, we have to show that there exist a $\text{c}\in(-1,2)$ such that f'(c) = 0
Now,
f(x) = (x2 - 1)(x - 2)
f'(x) = 2x(x - 2) + (x2 - 1)
= 2x2 - 4 + x2 - 1
f'(x) = 3x2 - 5
Now,
f'(c) = 0
⇒ 3x2 - 5 = 0
$\Rightarrow\text{x}=-\sqrt{\frac{5}{3}}$ or $\text{x}=\sqrt{\frac{5}{3}}\in(-1,2)$
Thus, Rolle's theorem is verified.
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$\vec{\text{r}}=\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}+\lambda\big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\big)$ and, $\vec{\text{r}}=3\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}+\mu\big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\big)$
$\vec{\text{c}}=\frac{1}{7}\big(6\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big),\hat{\text{i}},\hat{\text{j}},\hat{\text{k}}$
being a right handed orthogonal system of unit vector in spece, show that $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ is also another system.