Let [f:R R] be a continuous function defined by [ f( x ) = 1 e^x + 2e^ - x]

Question

Let $$f:R \to R$$ be a continuous function defined by
$$ f\left( x \right) = {1 \over {{e^x} + 2{e^{ - x}}}}$$

Accepted Answer

$$f\left( x \right) = {1 \over {{e^x} + 2{e^{ - x}}}} = {{{e^x}} \over {{e^{2x}} + 2}}$$

$$f'\left( x \right) = {{\left( {{e^{2x}} + 2} \right)e{}^x - 2{e^{2x}}.{e^x}} \over {{{\left( {{e^{2x}} + 2} \right)}^2}}}$$

$$f'\left( x \right) = 0 \Rightarrow {e^{2x}} + 2 = 2{e^{2x}}$$

$${e^{2x}} = 2 \Rightarrow {e^x} = \sqrt 2 $$

maximum $$f\left( x \right) = {{\sqrt 2 } \over 4} = {1 \over {2\sqrt 2 }}$$

$$0 < f\left( x \right) \le {1 \over {2\sqrt 2 }}\,\,\forall x \in R$$

Since $$0 < {1 \over 3} < {1 \over {2\sqrt 2 }} \Rightarrow $$ for some $$c \in R$$

$$f\left( c \right) = {1 \over 3}$$

Solutions · 1