Fx - 2x-1- x-0 2x-1- x-0 at x-0

When x=0+h (h is a very small real(+)ve number) x gt;0, f( x )=2x+1=2h+1=1 When x=0 h x lt;0 f( x )=2x 1= 2h 1= 1 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Critical Point

fx = 2x-1, ...

Question

$f (x) = {\begin{matrix} 2 x - 1, & x < 0 2 x + 1, & x \geq 0 \end{matrix}$ at $x = 0$

Open in App

Solution

Suggest Corrections

Similar questions

f (x) = \{\begin{array}{l} 2 x - 1, & x < 0 \\ 2 x + 1, & x \geq 0 \end{array} at x = 0 .

f (x) = {\begin{matrix} 3, x < 0 2 x + 1, x \geq 0 \end{matrix}

k

f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x} & i f - 1 \leq x < 0 \frac{2 x + 1}{x - 1} & i f 0 \leq x < 1 \end{matrix} ⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭

x = 0

f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{\sqrt{1 + p x} - \sqrt{1 - p x}}{x}, & i f - 1 \leq x < 0 \frac{2 x + 2}{x - 2}, & i f 0 \leq x < 1 \end{matrix}

x = 0

f (x) = \{\begin{cases} |x| \cos (\frac{1}{x}), & x \neq 0 \\ 0, & x = 0 \end{cases} at x = 0

f (x) = \{\begin{array}{l} x^{2} \sin (\frac{1}{x}), & x \neq 0 \\ 0, & x = 0 \end{array} at x = 0

f (x) = \{\begin{array}{l} (x - a) \sin (\frac{1}{x - a}), & x \neq a \\ 0, & x = a \end{array} at x = a

f (x) = \{\begin{array}{l} \frac{e^{x} - 1}{\log (1 + 2 x)}, if & x \neq a \\ 7, if & x = 0 \end{array} at x = 0

f (x) = \{\begin{array}{l} \frac{1 - x^{n}}{1 - x}, & x \neq 1 \\ n - 1, & x = 1 \end{array} n \in N at x = 1

f (x) = \{\begin{cases} \frac{|x^{2} - 1|}{x - 1}, for & x \neq 1 \\ 2, for & x = 1 \end{cases} at x = 1

f (x) = \{\begin{array}{l} \frac{2 |x| + x^{2}}{x}, & x \neq 0 \\ 0, & x = 0 \end{array} at x = 0

Join BYJU'S Learning Program