f(x) = x3 - 5x2 - 3x
Since, polynomial function is everywhere continuous and differentiable.
Therefore, f(x) is continuous on 1, 3 and differentiable on 1, 3
Thus, both the conditions of Lagrange's theorem is satisfied.
Concequently, there exist some $\text{c}\in1,3$ such that
$\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}=\frac{\text{f}(3)-\text{f}(1)}{2}$
Now, f(x) = x3 - 5x2 - 3x
f'(x) = 3x2 - 10x - 3
⇒ f(3) = -27
⇒ f(1) = -7
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(3)-\text{f}(1)}{2}$
$\Rightarrow3\text{x}^2-10\text{x}-3=\frac{-20}{2}$
$\Rightarrow3\text{x}^2-10\text{x}+7=0$
$\Rightarrow\text{x}=1,\frac{7}{3}$
Thus, $\text{c}=\frac{7}{3}\in(1,3)$ such that $\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}$
Hence, Lagrange's theorem is verified.
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| Bag | Number of White balls | Number of Black balls | Number of Red balls |
| I | 1 | 2 | 3 |
| II | 2 | 1 | 1 |
| III | 4 | 3 | 2 |
A bag is chosen at random and two balls are drawn from it. They happen to be white and red. What is the probability that they came from the III bag?
Find the minimum value of 3x + 5y subject to the constraints:
$-2\text{x}+\text{y}\leq4,\text{x}+\text{y}\geq3,$ $\text{x}-2\text{y}\leq2,\text{x},\text{y}\geq0.$