Answer: 943.26
Step-by-step explanation:
9.4326
x 100
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943.26
Hmm, the 2nd derivitve is good for finding concavity
let's find the max and min points
that is where the first derivitive is equal to 0
remember the difference quotient
so
f'(x)=(x^2-2x)/(x^2-2x+1)
find where it equals 0
set numerator equal to 0
0=x^2-2x
0=x(x-2)
0=x
0=x-2
2=x
so at 0 and 2 are the min and max
find if the signs go from negative to positive (min) or from positive to negative (max) at those points
f'(-1)>0
f'(1.5)<0
f'(3)>0
so at x=0, the sign go from positive to negative (local maximum)
at x=2, the sign go from negative to positive (local minimum)
we can take the 2nd derivitive to see the inflection points
f''(x)=2/((x-1)^3)
where does it equal 0?
it doesn't
so no inflection point
but, we can test it at x=0 and x=2
at x=0, we get f''(0)<0 so it is concave down. that means that x=0 being a max makes sense
at x=2, we get f''(2)>0 so it is concave up. that means that x=2 being a max make sense
local max is at x=0 (the point (0,0))
local min is at x=2 (the point (2,4))
It should be 24 because all degrees in angles add up to 180 so you add 49 & 107 get 156 subtract and boom!
The variance of a constant is zero. Rule 2. Adding a constant value, c, to a random variable does not change the variance, because the expectation (mean) increases by the same amount. ... Multiplying a random variable by a constant increases the variance by the square of the constant.
![2^{ \frac{3}{5}}= \sqrt[5]{2^3}](https://tex.z-dn.net/?f=%202%5E%7B%20%5Cfrac%7B3%7D%7B5%7D%7D%3D%20%20%5Csqrt%5B5%5D%7B2%5E3%7D%20%20)
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