If f x - e x -a for x-0 x-3 for x- 0 - is differentiable at x-0 then - a- is equal to-

The correct option is C 4 Given f(x) is differentiable at x=0 ⇒ f(x) is continuous at x=0 Now, f(0)=0 3= 3 LHL=lim _ x→ 0 ...

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Derivative from First Principle

if f x = e ...

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if $f (x) = {\begin{matrix} e^{x} + a f o r x < 0 x - 3 f o r x \geq 0 \end{matrix}$ is differentiable at $x = 0$
then ' $a$ ' is equal to?

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f (x) = {\begin{matrix} e^{x} + a x & x < 0 b (x - 1)^{2} & x \geq 0 \end{matrix}

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g (t) = {lim}_{x \to 0} {(1 + a tan x)}^{t / x}

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f (x) = {\begin{matrix} - 2, & - 3 \leq x \leq 0 x - 2, & x < x \leq 3 \end{matrix}

g (x) = f (| x |) + | f (x) |

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f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} x ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{{a e}^{\frac{1}{| x |}} + {3. e}^{\frac{- 1}{x}}}{(a + 2) e^{\frac{1}{| x |}} - e^{\frac{- 1}{x}}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 0 \end{matrix}, \begin{matrix} x \neq 0 x = 0 \end{matrix}

x = 0

[a] =

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{sin (α + 2) x + sin x}{x}, & x < 0 b, & x = 0 \frac{(x + 3 x^{2})^{\frac{1}{3}} - x^{\frac{1}{3}}}{x^{\frac{4}{3}}}, & x > 0 \end{matrix}

x = 0

a + 2 b

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