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Answer:
f'(x) > 0 on 
and f'(x)<0 on
Step-by-step explanation:
1) To find and interval where any given function is increasing, the first derivative of its function must be greater than zero:
To find its decreasing interval :
2) Then let's find the critical point of this function:
![f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[6-2^{2x}]=\frac{\mathrm{d} }{\mathrm{d}x}[6]-\frac{\mathrm{d}}{\mathrm{d}x}[2^{2x}]=0-[ln(2)*2^{2x}*\frac{\mathrm{d}}{\mathrm{d}x}[2x]=-ln(2)*2^{2x}*2=-ln2*2^{2x+1\Rightarrow }f'(x)=-ln(2)*2^{2x}*2\\-ln(2)*2^{2x+1}=-2x^{2x}(ln(x)+1)=0](https://tex.z-dn.net/?f=f%27%28x%29%3D%5Cfrac%7B%5Cmathrm%7Bd%7D%20%7D%7B%5Cmathrm%7Bd%7D%20x%7D%5B6-2%5E%7B2x%7D%5D%3D%5Cfrac%7B%5Cmathrm%7Bd%7D%20%7D%7B%5Cmathrm%7Bd%7Dx%7D%5B6%5D-%5Cfrac%7B%5Cmathrm%7Bd%7D%7D%7B%5Cmathrm%7Bd%7Dx%7D%5B2%5E%7B2x%7D%5D%3D0-%5Bln%282%29%2A2%5E%7B2x%7D%2A%5Cfrac%7B%5Cmathrm%7Bd%7D%7D%7B%5Cmathrm%7Bd%7Dx%7D%5B2x%5D%3D-ln%282%29%2A2%5E%7B2x%7D%2A2%3D-ln2%2A2%5E%7B2x%2B1%5CRightarrow%20%7Df%27%28x%29%3D-ln%282%29%2A2%5E%7B2x%7D%2A2%5C%5C-ln%282%29%2A2%5E%7B2x%2B1%7D%3D-2x%5E%7B2x%7D%28ln%28x%29%2B1%29%3D0)
2.2 Solving for x this equation, this will lead us to one critical point since x' is not defined for Real set, and x''
≈0.37 for e≈2.72
3) Finally, check it out the critical point, i.e. f'(x) >0 and below f'(x)<0.
First question:
4(3x-5) + (2x -2)(4 + 3x-5)
Simplify -----> 12x - 20 + (2x-2)(3x-1)
Multiply the second part -----> 12x - 20 + 6x^2 -8x + 2
Combine like terms --------> 6x^2 + 4x - 18
Each side of the addition sign represents one of the rectangles above which is then added to get the area of the entire figure.
Second question:
The average is the summation of each data value divided by the number of data values. So it would be
(4 + 5 + 9)/ 3
Simplify ------> 18/3
Divide --------> 6
See, when you are traveling, you might get lost. if so, just remember "I need to go between _ and _ miles" estimating and rounding to the nearest number with 0 wiil help.
[tex]<span>1\dfrac{1}{7}=\dfrac{1\cdot7+1}{7}=\dfrac{8}{7}\\\\1\dfrac{1}{7}:\dfrac{1}{9}=\dfrac{8}{7}:\dfrac{1}{9}=\dfrac{8}{7}\cdot\dfrac{9}{1}=\dfrac{72}{7}[tex]
Answer: 1 1/7 : 1/9 = 72 : 7
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