The average rate of change (AROC) of a function f(x) on an interval [a, b] is equal to the slope of the secant line to the graph of f(x) that passes through (a, f(a)) and (b, f(b)), a.k.a. the difference quotient given by
![f_{\mathrm{AROC}[a,b]} = \dfrac{f(b)-f(a)}{b-a}](https://tex.z-dn.net/?f=f_%7B%5Cmathrm%7BAROC%7D%5Ba%2Cb%5D%7D%20%3D%20%5Cdfrac%7Bf%28b%29-f%28a%29%7D%7Bb-a%7D)
So for f(x) = x² on [1, 5], the AROC of f is
![f_{\mathrm{AROC}[1,5]} = \dfrac{5^2-1^2}{5-1} = \dfrac{24}4 = \boxed{6}](https://tex.z-dn.net/?f=f_%7B%5Cmathrm%7BAROC%7D%5B1%2C5%5D%7D%20%3D%20%5Cdfrac%7B5%5E2-1%5E2%7D%7B5-1%7D%20%3D%20%5Cdfrac%7B24%7D4%20%3D%20%5Cboxed%7B6%7D)
Answer:
a ≈ 21.8
Step-by-step explanation:
We require the third angle in the triangle
Subtract the sum of the 2 angles from 180
third angle = 180° - (75 + 31.8)° = 180° - 106.8° = 73.2°
Using the Sine rule, that is
= 
( cross- multiply )
a × sin75° = 22 × sin73.2° ( divide both sides by sin75° )
a = 
≈ 21.8 ( to the nearest tenth )
Reflections and glide reflections both give you a transformation where the preimage and image have opposite orientations.
The roots of the polynomial <span><span>x^3 </span>− 2<span>x^2 </span>− 4x + 2</span> are:
<span><span>x1 </span>= 0.42801</span>
<span><span>x2 </span>= −1.51414</span>
<span><span>x3 </span>= 3.08613</span>
x1 and x2 are in the desired interval [-2, 2]
f'(x) = 3x^2 - 4x - 4
so we have:
3x^2 - 4x - 4 = 0
<span>x = ( 4 +- </span><span>√(16 + 48) </span>)/6
x_1 = -4/6 = -0.66
x_ 2 = 2
According to Rolle's theorem, we have one point in between:
x1 = 0.42801 and x2 = −1.51414
where f'(x) = 0, and that is <span>x_1 = -0.66</span>
so we see that Rolle's theorem holds in our function.
