Y=1/x-2+1 the domain and range of this function
1 answer:
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Answer:
f(t) = -t + 21
g(t) = 18*e^( - t / 12 + 1/4 )
Step-by-step explanation:
Given:
- The graphs for the similar question is attached.
- The same graph would be used as reference but with different coordinates for point of intersection of f(t) and g(t) @ ( 3 , 18 ) & ( 15 , 6 ).
Find:
- The formulas for functions f(t) and g(t).
Solution:
- First we will determine f(t) the blue graph which is a "linear" function. The general equation for the linear function is given as:
f(t) = m*t + c
Where, m: is the gradient ( constant )
c: The f(t) intercept. ( constant )
- The gradient m can be determined by the given points that lie on the graph:
m = ( f(t2) - f(t1) ) / ( t2 - t1 )
m = ( 6 - 18 ) / ( 15 - 3 )
m = -12 / 12 = -1
- The constant c can be evaluated by using any one point and m substituted back into the linear expression as follows:
f(t) = -t + c
18 = -(3) + c
c = 21
- The function f(t) is as follows:
f(t) = -t + 21
- The general expression for an exponential function can be written as:
g(t) = a*e^(b*t)
Where, a and b are constants to be evaluated.
- We will develop two expressions for g(t) using two given points that lie on the curve as follows:
18 = a*e^(3*b)
6 = a*e^(15*b)
- Divide the two expressions we have:
3 = e^( 3b - 15b )
Ln(3) = -12*b
b = - Ln(3) / 12
- Then the expression 1 becomes:
18 = a*e^( - Ln(3)*3 / 12)
18 = 3*a*e^(-1/4)
6 = a / e^(0.25)
a = 6*e^( 1 / 4 )
- The function g(t) can be expressed as:
g(t) = 18*e^( - t / 12 + 1/4 )
