H(x) = f(x)/g(x)
Use the quotient rule.
h '(x) = [g(x) f '(x) - f(x) g '(x)] / [g(x)]^2
=> h '(1) = [g(1) f '(1) - f(1) g '(1)] / [(g(1) ] ^2
h '(1) = [ 3*(-4) - 4*(-3)] / (3)^2 = [-12 + 12] / 9 = 0
Answer: 0
let's recall that on the IV Quadrant the x/cosine is positive and the y/sine is negative, and of course the hypotenuse is just a radius unit and therefore never negative.

![\bf \pm\sqrt{15^2-(-7)^2}=a\implies \pm\sqrt{176}=a\implies \stackrel{\textit{IV Quadrant}}{+\sqrt{176}=a} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill cos(\theta )=\cfrac{\stackrel{adjacent}{\sqrt{176}}}{\stackrel{hypotenuse}{15}}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cpm%5Csqrt%7B15%5E2-%28-7%29%5E2%7D%3Da%5Cimplies%20%5Cpm%5Csqrt%7B176%7D%3Da%5Cimplies%20%5Cstackrel%7B%5Ctextit%7BIV%20Quadrant%7D%7D%7B%2B%5Csqrt%7B176%7D%3Da%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20cos%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Badjacent%7D%7B%5Csqrt%7B176%7D%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B15%7D%7D~%5Chfill)
-14+6b+7-2b=1+5b
step 1: simplify the equation
-7+4b=1+5b
step 2: collect your variables
-7=1+b
step 3: collect your whole numbers
-8=b
Answer:
<h2>
X=1.33</h2>
solution,

Hope this helps.
Good luck on your assignment.m