Be the exponential function given by f(x) = (1/2)^(x^2-8x+15), determine the greatest value that f(x) can take.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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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` .

lilahrose8248's profile pic

lilahrose8248 | Student, Grade 9 | (Level 1) eNoter

Posted on

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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