Let f1-2 and f-x - 4-2 for 1- x - 6- The smallest possible value of f6 is

The correct option is B 19 By Lagrange's mean value theorem, f(6) f(1)6 1=f'(c) f(6)+25≥ 4.2 f(6)+2≥ 21 f(6)≥ 19 ...

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Standard XII

Mathematics

Sufficient Condition for an Extrema

Let f1=-2 a...

Question

Let $f (1) = - 2$ and $f^{'} (x) \geq 4.2$ for $1 \leq x \leq 6$ . The smallest possible value of $f (6)$ is

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Solution

The correct option is B $19$
By Lagrange's mean value theorem,
$\frac{f (6) - f (1)}{6 - 1} = f^{'} (c)$
$\frac{f (6) + 2}{5} \geq 4.2$
$f (6) + 2 \geq 21$
$f (6) \geq 19$

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Let f(1)=−2 and f′(x)≥4.2 for 1≤x≤6. The smallest possible value of f(6) is

The correct option is B 19By Lagrange's mean value theorem,f(6)−f(1)6−1=f′(c)f(6)+25≥4.2f(6)+2≥21f(6)≥19

Let $f (1) = - 2$ and $f^{'} (x) \geq 4.2$ for $1 \leq x \leq 6$ . The smallest possible value of $f (6)$ is

The correct option is B $19$
By Lagrange's mean value theorem,
$\frac{f (6) - f (1)}{6 - 1} = f^{'} (c)$
$\frac{f (6) + 2}{5} \geq 4.2$
$f (6) + 2 \geq 21$
$f (6) \geq 19$