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You should remember that the use of derivatives may help you in finding the maximum or minimum values a function can take, hence, you need to determine `f'(x)` and you need to solve for x `f'(x) = 0` such that:
`f(x) = (1/2)^(x^2-8x+15) => ln f(x) = (x^2-8x+15)*ln(1/2)`
Differentiating both sides yields:
`1/(f(x))*f'(x) = (2x-8)*ln(1/2)`
`f'(x) = (2x-8)*ln(1/2)*((1/2)^(x^2-8x+15))`
You need to solve for x the equation `f'(x)=0` such that:
`(2x-8)*ln(1/2)*((1/2)^(x^2-8x+15)) = 0`
Since `((1/2)^(x^2-8x+15)) > 0` for `x in R` , hence `2x-8 = 0` such that:
`2x - 8 = 0 => 2x = 8 => x = 4`
You need to evaluate f(4) such that:
`f(4) = (1/2)^(4^2-32+15) => f(4) = (1/2)^(31-32) => f(4) = (1/2)^(-1) = 1/(1/2) => f(4) = 2` .
Hence, evaluating the maximum value the function can take, under the given conditions, yields `f(4) = 2` .
I'm not really sure what you're asking in the question, especially when you say what f(x) can take.
But if mean what is the value of x that will maximize f(x), I can answer that.
All you need to do is plug that equation into a graphing calculator and graph it. Then, you use the maximum key, and it gives you the maximum of the graph.
You would get (4.0000015, 2), meaning the x value that would maximize the function is approximately 4.
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