1. ## Derivative and graphing!!!

The graphs of the function (given in blue) and (given in red) are plotted above. Find each of the following:
= ?
= ?

2. If

find the values of and that make differentiable everywhere. m= ? and b=?

2. Hello, Kayla_N!

The graphs of the function $\displaystyle f$ (in blue) and $\displaystyle g$ (in red) are plotted above.

Find the following: .$\displaystyle (1)\;(fg)'(1)\qquad (2)\;\left(\frac{f}{g}\right)\!'(1)$
We have piecewise functions . . .

$\displaystyle f(x) \;=\;\bigg\{ \begin{array}{cccc}\frac{3}{2}x & & 0 < x < 3 \\ \frac{9}{2}-x & & x > 3 \end{array}$

$\displaystyle g(x) \;=\;\bigg\{\begin{array}{cccc}4-x & & x < 2 \\ \frac{1}{2}x + 2 & & x > 2 \end{array}$

If $\displaystyle x = 1$, we have: .$\displaystyle \begin{array}{ccc}f(x) \;=\; \frac{3}{2}x \\ g(x) \;=\; 4 - x \end{array}$

$\displaystyle (1)\;\;(fg) \;=\;\tfrac{3}{2}x(4 - x) \;=\;6x - \tfrac{3}{2}x^2$

Then:. . $\displaystyle (fg)' \;=\;6 - 3x$

Therefore: .$\displaystyle (fg)'(1) \;=\;6-3(1) \;=\;\boxed{3}$

$\displaystyle (2)\;\;\left(\frac{f}{g}\right) \;=\;\frac{\frac{3}{2}x}{4-x} \;=\;\frac{3}{2}\cdot\frac{x}{4-x}$

Then: .$\displaystyle \left(\frac{f}{g}\right)\!' \;=\;\frac{3}{2}\cdot\frac{(4-x)\!\cdot\!1 - x(-1)}{(4-x)^2} \;=\;\frac{6}{(4-x)^2}$

Therefore: .$\displaystyle \left(\frac{f}{g}\right)\!'(1) \;=\;\frac{6}{(4-1)^2} \;=\;\frac{6}{9} \;=\;\boxed{\frac{2}{3}}$

3. Thanks for explaining steps by steps. Thats really helped and i understand it now.