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Chain Rule of Differentiation

lf fx=|x2-5...

Question

lf $f (x) = | x^{2} - 5 x + 6 |$ then $f^{'} (2^{+})$ is

A

0

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B

1

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C

- 1

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D

2

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Solution

The correct option is B $1$
Given, $f (x) = | (x - 2) (x - 3) | = | x - 2 | | x - 3 |$
$f (x) = (x - 2) (x - 3)$ , $x > 3$ and $x < 2$
$= - (x - 2) (x - 3)$ , $2 < x < 3$
$f^{'} (x) ∣_{x = 2^{+}} = \frac{d}{d x} [- (x - 2) (x - 3)] = - (x - 3) - (x - 2)$
$= - (- 1) - 0$
$= 1$

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Similar questions

f (x) = \frac{x^{2}}{2}

0 \leq x \leq 1

f (x) = 2 x^{2} - 3 x + \frac{3}{2}

1 \leq x \leq 2

then the function

f^{''}

f (x) = \frac{x^{2}}{2}

0 \leq x \leq 1

f (x) = 2 x^{2} - 3 x + (3 / 2)

1 \leq x \leq 2

then the function

f^{''} (x)

f (x) = \frac{x^{3}}{x^{2} - 1}

f^{^{'''}} (0) =

f (x) = x - x^{2} + x^{3} - x^{4} + \dots .

| x | < 1

f^{^{''}} (0) =

f (x) = \frac{1}{x^{2}} \int_{4}^{x} (4 t^{2} - 2 f^{'} (t)) d t

f^{'} (4) =

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