What is the one-sided limit of the derivative at x = 2 for the given function?

[cbarker1]

Finding the one-sided derivatives for f, when c=2.f(x)={\begin{array}{cc}x/2+1,&\mbox{ if }
x< 2\\\sqrt{2x}, & \mbox{ if } x\geq 2\end{array}

Right-side derivative is 1/2. But I have trouble with the left-side.

Thank You

Cbarker1

 

[Fermat1]

Finding the one-sided derivatives for f, when c=2.f(x)={\begin{array}{cc}x/2+1,&\mbox{ if }
x< 2\\\sqrt{2x}, & \mbox{ if } x\geq 2\end{array}

Right-side derivative is 1/2. But I have trouble with the left-side.

Thank You

Cbarker1

Try again, there is no difficulty. You should find the left hand limit is 1/2

 

[MarkFL]

Finding the one-sided derivatives for f, when c=2.f(x)={\begin{array}{cc}x/2+1,&\mbox{ if }
x< 2\\\sqrt{2x}, & \mbox{ if } x\geq 2\end{array}

Right-side derivative is 1/2. But I have trouble with the left-side.

Thank You

Cbarker1

Can you show us what you tried for the left side? Are you differentiating from first principles (the difference quotient limit) or using standard rules of differentiation?

Incidentally, if you wish to use $\LaTeX$ to write a piecewise function, use the begin/end environment as follows:

[noparsetex]\(\displaystyle f(x)=\begin{cases}\dfrac{x}{2}+1 & x<2 \\&\\ \sqrt{2x} & x\qe 2 \\\end{cases}\)[/noparsetex]

to get:

\(\displaystyle f(x)=\begin{cases}\dfrac{x}{2}+1 & x<2 \\&\\ \sqrt{2x} & x\ge 2 \\\end{cases}\)

 

[cbarker1]

Thank you for your tip on Latex. I have written the Attempt on the paper.View attachment 2451

 

[MarkFL]

First, let's approach $x=2$ from the left:

\(\displaystyle \lim_{x\to2^{-}}\left(f'(x)\right)\)

For $x<2$, we are given:

\(\displaystyle f(x)=\frac{x}{2}+1\)

Hence:

\(\displaystyle f'(x)\equiv\lim_{h\to0}\left(\frac{f(x+h)-f(x)}{(x+h)-x}\right)=\lim_{h\to0}\left(\frac{\left(\dfrac{x+h}{2}+1\right)-\left(\dfrac{x}{2}+1\right)}{h}\right)\)

Can you complete the simplification to obtain the value of the derivative, and then using that, find the one-sided limit?

Then, do the same for the right-side. If you get stuck, post your progress, and we will help. :D