If f(x)=(2 k+1) x-3-k e^{-x}+2 e^x is monoton | vedclass

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(B) Given,$f(x)=(2 k+1) x-3-k e^{-x}+2 e^x$.Since $f(x)$ is monotonically increasing for all $x \in R$,its derivative $f'(x)$ must be greater than or

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(B) Given,$f(x)=(2 k+1) x-3-k e^{-x}+2 e^x$.Since $f(x)$ is monotonically increasing for all $x \in R$,its derivative $f'(x)$ must be greater than or equal to $0$ for all $x \in R$.$f'(x) = (2k+1) + k e^{-x} + 2 e^x \geq 0$.Multiplying by $e^x$ (which is always positive),we get:$(2k+1)e^x + k + 2e^{2x} \geq 0$.Rearranging the terms:$2e^{2x} + (2k+1)e^x + k \geq 0$.Factoring the quadratic expression in terms of $e^x$:$2e^{2x} + 2ke^x + e^x + k \geq 0$.$2e^x(e^x + k) + 1(e^x + k) \geq 0$.$(2e^x + 1)(e^x + k) \geq 0$.Since $2e^x + 1 > 0$ for all $x \in R$,we must have $e^x + k \geq 0$ for all $x \in R$.This implies $k \geq -e^x$ for all $x \in R$.As $x 	o -\infty$,$e^x 	o 0$,so $k \geq 0$.Thus,the least value of $k$ is $0$.

If $f(x)=(2 k+1) x-3-k e^{-x}+2 e^x$ is monotonically increasing for all $x \in R$,then the least value of $k$ is

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