range of (4^x)+(2^x)+1 is

Hi  Kshitij,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=4^{\mathrm{x}}+2^{\mathrm{x}}+1 \\ \Rightarrow \mathrm{f}\text{'}\left(\mathrm{x}\right)=4^{\mathrm{x}}.\ln \left(4\right)+2^{\mathrm{x}}.\ln \left(2\right) \\ \mathrm{Now} \mathrm{since} {\mathrm{a}}^{\mathrm{x}} \mathrm{is} \mathrm{always} \mathrm{positive}, \\ \Rightarrow \mathrm{f}\text{'}\left(\mathrm{x}\right)>0 \mathrm{for} \mathrm{all} \mathrm{x}\in \mathrm{R} \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right) \mathrm{is} \mathrm{an} \mathrm{increasing} \mathrm{function}. \mathrm{Hence} \mathrm{it} \mathrm{will} \mathrm{have} \mathrm{its} \mathrm{minimum} \mathrm{value} \mathrm{when} \mathrm{x}\to -\infty \mathrm{and} \mathrm{maximum} \mathrm{value} \mathrm{when} \mathrm{x}\to \infty \\ \mathrm{Now} \underset{\mathrm{x}\to -\infty }{\lim }{\mathrm{a}}^{\mathrm{x}}=0 \\ \Rightarrow \underset{\mathrm{x}\to -\infty }{\lim }{4}^{\mathrm{x}}+2^{\mathrm{x}}=0 \\ \Rightarrow \mathrm{Hence} \min \mathrm{value} \mathrm{of} \mathrm{f}\left(\mathrm{x}\right)=4^{\mathrm{x}}+2^{\mathrm{x}}+1 \mathrm{will} \mathrm{tend} \mathrm{towards} 1 \mathrm{and} \mathrm{it} \mathrm{is} \mathrm{clear} \mathrm{that} \mathrm{maximum} \mathrm{value} \mathrm{will} \mathrm{tend} \mathrm{towards} \infty . \\ \mathbf{Hence}\mathbf{ }\mathbf{range}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{f}\left(\mathbf{x}\right)\mathbf{=}{\mathbf{4}}^{\mathbf{x}}\mathbf{+}{\mathbf{2}}^{\mathbf{x}}\mathbf{+}\mathbf{1}\mathbf{ }\mathbf{is}\mathbf{ }\left(\mathbf{1}\mathbf{,}\mathbf{ }\mathbf{\infty }\right) \end{gathered}$$

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