If f-x- - -x-2 - 5x - 6- then f-x- equals2x -5 for 2 x 35 -2x for 2 x 32x -5 for x 25 - 2x for x 3

The correct options are B 5 2x for 2 x 3 C 2x 5 for x 2We know that, |x^2 nbsp; 5x + 6| = |(x 3)(x 2)| So the function will be, f(x) = x^2 5x +6 ...

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First Principle of Differentiation

If f(x) = |x^...

Question

If $f (x) = | x^{2} - 5 x + 6 |$ , then $f^{'} (x)$ equals

2 x - 5

for

2 < x < 3

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5 - 2 x

for

2 < x < 3

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2 x - 5

for

x < 2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

5 - 2 x

for

x < 3

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I = \int \frac{x^{3}}{\sqrt{x^{2} + 2 x + 5}} d x = \frac{1}{6} f (x) - \frac{A}{120} g (x) + C

f (x) = (2 x^{2} - 5 x - 36) \sqrt{x^{2} + 2 x + 5}, g (x) = log (x + 1 + \sqrt{x^{2} + 2 x + 5})

\frac{x^{2} - 7 x + 10}{x^{2} - 5 x - 14} \times \frac{x^{3} + 8}{x^{3} - 8} \times \frac{x^{2} + 2 x + 4}{x^{2} - 2 x + 4} \times \frac{x - 7}{x - 5} = ?

$If f (x) = (\begin{matrix} x^{3} + x^{2}, f o r 0 \leq x \leq 2 x + 2, f o r 2 \leq x \leq 4 \end{matrix}$
then the odd extension of f(x) would be -

lim x \to 2 \frac{2 x + 5}{8 - x^{3}}

(x^{3} - 4 x^{2} + 2 x + 5) \div (x - 2)

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