You first switch f(x) with x in this equation
<span>x=<span>1<span>9f(x)−2</span></span></span>
Then you solve for f(x).
<span>9f(x)−2=<span>1x</span></span>
<span>9f(x)=<span>1x</span>+2</span>
<span>f(x)=<span>1<span>/9x</span></span>+<span>2/9</span></span>
Then you simply repalace f(x) with f^-1(x)
It will look like this
<span><span>f−</span>1(x)=<span>1<span>/9x</span></span>+<span>2/<span>9
Hope i could help
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Answer:
Domain → (-∞, -1)∪(-1, ∞)
Range → (-∞, ∞)
Step-by-step explanation:
Given function is,
This function is not defined when the denominator is zero.
2x + 2 = 0
2x = -2
x = -1
Therefore, x = -1 is not in the domain of this function.
And the domain will be,
(-∞, -1)∪(-1, ∞)
For all values of x (except x = -1) we get some output values (value of 
).
Therefore, range of the function will be (-∞, ∞).
Answer:
D) 7.5%
Step-by-step explanation:
Using dividend growth model,
Price, P = D / (r - g)
where, D = Dividend, r = cost of equity = 10% , and g= growth rate
we know that
=> r - g = D / P = Dividend yield = 2.5%
=> Growth rate, g = 10% - 2.5% = 7.5%
therefore the constant growth rate in dividends is closest to= 7.5%
Answer:
L = ∫₀²ᵖⁱ √((1 − sin t)² + (1 − cos t)²) dt
Step-by-step explanation:
Arc length of a parametric curve is:
L = ∫ₐᵇ √((dx/dt)² + (dy/dt)²) dt
x = t + cos t, dx/dt = 1 − sin t
y = t − sin t, dy/dt = 1 − cos t
L = ∫₀²ᵖⁱ √((1 − sin t)² + (1 − cos t)²) dt
Or, if you wish to simplify:
L = ∫₀²ᵖⁱ √(1 − 2 sin t + sin²t + 1 − 2 cos t + cos²t) dt
L = ∫₀²ᵖⁱ √(3 − 2 sin t − 2 cos t) dt
