A linear function and an exponential function are graphed below. Find possible formulas for the functions f(t), in blue, and g(t
), in red, that go through the points (3,18) and (15,6).
1 answer:
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 )
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