The minimum value of f(x) = a^{a^{x}} + a^{1- | vedclass

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(B) Given the function $f(x) = a^{a^{x}} + a^{1-a^{x}}$.We can rewrite the second term as $a^{1-a^{x}} = \frac{a}{a^{a^{x}}}$.So,$f(x) = a^{a^{x}} + \

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$2\sqrt{a}$

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(B) Given the function $f(x) = a^{a^{x}} + a^{1-a^{x}}$.We can rewrite the second term as $a^{1-a^{x}} = \frac{a}{a^{a^{x}}}$.So,$f(x) = a^{a^{x}} + \frac{a}{a^{a^{x}}}$.Since $a > 0$,both $a^{a^{x}}$ and $\frac{a}{a^{a^{x}}}$ are positive real numbers.By the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality,for any two positive numbers $u$ and $v$,$u + v \geq 2\sqrt{uv}$.Let $u = a^{a^{x}}$ and $v = \frac{a}{a^{a^{x}}}$.Then $f(x) = u + v \geq 2\sqrt{u \cdot v} = 2\sqrt{a^{a^{x}} \cdot \frac{a}{a^{a^{x}}}} = 2\sqrt{a}$.Thus,the minimum value is $2\sqrt{a}$.

The minimum value of $f(x) = a^{a^{x}} + a^{1-a^{x}}$,where $a, x \in R$ and $a > 0$,is equal to ..... .

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