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.
538
-247
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8-7=1
Cross out 5 and make it 4. Make 3 into 13. 13-4=9
4-2=2
The answer is: 291
Yes, they are equivalent
6+7-5=8. -5+10+3=8
Answer:
(–5, –7)
Step-by-step explanation:
From the question given above, the following data were obtained:
Slope = 9/5
Coordinate 1 = (–10, –16)
x₁ = –10
y₁ = –16
Coordinate 2 = (x₂, y₂)
Next, we shall determine the change in x and y coordinate. This can be obtained as follow:
Slope = change in y–coordinate / change in x–coordinate
Slope = Δy / Δx
Slope = 9/5
9/5 = Δy / Δx
Thus,
Δy = 9
Δx = 5
Next, we shall determine the second coordinates as follow:
Δy = y₂ – y₁
Δx = x₂ – x₁
For x–coordinate:
x₁ = –10
Δx = 5
Δx = x₂ – x₁
5 = x₂ – (–10)
5 = x₂ + 10
Collect like terms
x₂ = 5 – 10
x₂ = – 5
For y–coordinate:
y₁ = –16
Δy = 9
Δy = y₂ – y₁
9 = y₂ – (–16)
9 = y₂ + 16
Collect like terms
y₂ = 9 – 16
y₂ = – 7
Coordinate 2 = (x₂, y₂)
Coordinate 2 = (–5, –7)