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Monotonically Increasing Functions

For which val...

Question

For which value of $x$ , the function $f (x) = x^{2} - 2 x$ is decreasing

x > 1

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x > 2

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x < 1

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x < 2

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Solution

The correct option is B $x < 1$
$f (x) = x^{2} - 2 x$
$f^{'} (x) = 2 x - 2$
$f (x) i s d e c r e a sin g$
$i f f (x) < 0$
$2 x - 2 < 0$
$2 x < 2$
$x < 1$

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Similar questions

f (x)

[- 2, 2]

(a + b)

f (x) = \frac{sin a x}{x} - 2,

- 2 \leq x < 0

= 2 x + 1,

0 \leq x \leq 1

= 2 b \sqrt{x^{2} + 3 - 1},

1 < x \leq 2

f (x) = ⎧ ⎨ ⎩ \begin{matrix} a x - 2 & f o r & - 2 < x < - 1 - 1 & f o r & - 1 \leq x \leq 1 a + 2 (x - 1)^{2} & f o r & 1 < x < 2 \end{matrix}

f (x)

x = - 1

x = 1

x

f (x) = ⎧ ⎨ ⎩ \begin{matrix} (1 - x), x < 1 (1 - x) (2 - x), 1 \leq x \leq 2 (3 - x), x > 2 \end{matrix}

f (x) = \{\begin{matrix} 2 x + 1; & if x < 2 \\ k; & x = 2 \\ 3 x - 1; & x > 2 \end{matrix}

f (x) = x^{2} + 2 x + 2; x \geq - 1

= - x^{2} + 3 x + 1; x < - 1

f (f (- 2))

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