If f, of, xf(x) is an exponential function where f, of, minus, 3, equals, 24f(−3)=24 and f, of, 0, point, 5, equals, 68f(0.5)=68, then find the value of f, of, minus, 2, point, 5f(−2.5), to the nearest hundredth.
Since f(x) is an exponential function, it can be written in the form f(x) = a * b^x, where a and b are constants.
Given that f(-3) = 24, we have 24 = a * b^(-3).
Similarly, f(0.5) = 68, which gives us 68 = a * b^(0.5).
Dividing these two equations, we can eliminate the constant a:
(24/68) = (a * b^(-3))/(a * b^(0.5))
(24/68) = b^(-3-0.5)
(24/68) = b^(-3.5)
Taking the reciprocal of both sides, we get:
(68/24) = 1/(b^(-3.5))
(68/24) = b^(3.5)
Now, we have two equations with two unknowns, a and b:
24 = a * b^(-3)
(68/24) = b^(3.5)
To find the value of f(-2.5), we need to evaluate the expression f(-2.5) = a * b^(-2.5).
To solve for a and b, we can substitute the value of b from the second equation into the first equation:
24 = a * ((68/24)^(3.5))^(3/2)
Simplifying this equation will give us the values of a and b.
Calculating the equation, we get a ≈ 1.01588 and b ≈ 1.7393.
Now, we can find the value of f(-2.5):
f(-2.5) = 1.01588 * (1.7393)^(-2.5)
≈ 1.01588 * 0.468589
≈ 0.47721
Therefore, the value of f(-2.5) is approximately 0.48.