Question

(\ f(x) = \sqrt{x+1} \)

Answer 1

(/ f(x) = \sqrt{x+1} /)

Answer 2

/( f(x) = \sqrt{x+1} /)

Answer 3

\( f(x) = \sqrt{x+1} \)

Answer 4

```
\(\huge f(x) = \sqrt{x+1}\)
```
\(\huge f(x) = \sqrt{x+1}\)

Answer 5

Hello @LarsEighner !
U need help ?

Answer 6

\( {\displaystyle \color{red}{\lim_{h \rightarrow 0} { {f(x+h) - f(x)} \over h} }} \)

Answer 7

\({\huge\displaystyle \color{green}{\lim_{h \rightarrow 0} { {f(x+h) - f(x)} \over h} }}\)

Answer 8

\(\Huge{\overset{\frown}{\normalsize
\left(
\begin{matrix}
\Large\cdot \quad \cdot\\
\cdot\\
\huge \smile
\end{matrix}
\right)}}\normalsize
\\
\;/\quad \;\;\quad \backslash\)

Answer 9

@abhisar Thanks I think I'm getting it. I am a mathjax wizard but I just need to find the knobs hers.

Answer 10

oh i thought u need help

Answer 11

\(\huge\color{green}{ت}\)

Answer 12

"I just need to find the knobs hers"......sorry i didn't got this !

Answer 13

\( \begin{align} f^\prime (3) &= \lim_{ h \rightarrow0} {{f(3 + h) -f(3)}
\over { h}} \cr &= \lim_{ h \rightarrow0} {{(4+h)^{1\over 2} -2}
\over { h}} \cr &= \lim_{ h \rightarrow0} {{[(4+h)^{1\over 2} -2][(4+h)^{1\over 2} +2]}
\over { h[(4+h)^{1\over 2} +2]}} \cr &= \lim_{ h \rightarrow0} {{[(4+h)^{1\over 2}]^2 -4}
\over { h[(4+h)^{1\over 2} +2]}} \cr &= \lim_{ h \rightarrow0} {{(4+h) -4}
\over { h[(4+h)^{1\over 2} +2]}} \cr &= \lim_{ h \rightarrow0} {h
\over { h[(4+h)^{1\over 2} +2]}} \cr &= \lim_{ h \rightarrow0} {1
\over { (4+h)^{1\over 2} +2}} \cr &= {1
\over { (4+0)^{1\over 2} +2}} \cr f^\prime (3) &= {1
\over 4} \end{align} \)

Answer 14

hers = here

Answer 15

oh !