# Derivative and graphing!!!

• February 21st 2009, 11:28 AM
Kayla_N
Derivative and graphing!!!
• February 21st 2009, 01:41 PM
Soroban
Hello, Kayla_N!

Quote:

http://webwork1.math.utah.edu/webwor...ob19image1.png

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

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

We have piecewise functions . . .

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

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

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

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

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

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

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

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

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

• February 21st 2009, 04:09 PM
Kayla_N
Thanks for explaining steps by steps. Thats really helped and i understand it now.