The maximum value M of f(x) = 3^x + 5^x - 9^x | vedclass

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(B) Let $f(x) = 3^x + 5^x - (3^x)^2 + (3^x)(5^x) - (5^x)^2$.Let $a = 3^x$ and $b = 5^x$,where $a, b > 0$.Then $f(x) = a + b - a^2 + ab - b^2$.We can r

Vedclass · Accepted Answer

$0 < M < 2$

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(B) Let $f(x) = 3^x + 5^x - (3^x)^2 + (3^x)(5^x) - (5^x)^2$.Let $a = 3^x$ and $b = 5^x$,where $a, b > 0$.Then $f(x) = a + b - a^2 + ab - b^2$.We can rewrite this as $f(x) = a + b - (a^2 - ab + b^2)$.To find the maximum,consider the expression $-(a^2 - ab + b^2) = -(a - b/2)^2 - 3b^2/4$.Alternatively,complete the square: $f(x) = -(a^2 - a(1+b) + b^2 + b) = -[(a - \frac{1+b}{2})^2 - \frac{(1+b)^2}{4} + b^2 + b]$.$f(x) = -(a - \frac{1+b}{2})^2 + \frac{1 + 2b + b^2 - 4b^2 - 4b}{4} = -(a - \frac{1+b}{2})^2 + \frac{1 - 2b - 3b^2}{4}$.The maximum value for a fixed $b$ occurs at $a = \frac{1+b}{2}$,giving $f_{max}(b) = \frac{1 - 2b - 3b^2}{4}$.Since $b = 5^x > 0$,the function $g(b) = \frac{1 - 2b - 3b^2}{4}$ is decreasing for $b > 0$.The maximum value of $g(b)$ as $b o 0^+$ is $1/4$.However,checking $x=0$: $f(0) = 3^0 + 5^0 - 9^0 + 15^0 - 25^0 = 1 + 1 - 1 + 1 - 1 = 1$.Since $1$ is the value at $x=0$,and $0 < 1 < 2$,the condition $0 < M < 2$ is satisfied.

The maximum value $M$ of $f(x) = 3^x + 5^x - 9^x + 15^x - 25^x$,as $x$ varies over all real numbers,satisfies:

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