Question

Show that f(x) = (x − 1) e^x + 1 is an increasing function for all x > 0.

Answer 1

$$\begin{gathered} f\left(x\right)=\left(x-1\right)e^x+1 \\ f\text{'}\left(x\right)=\left(x-1\right)e^x+e^x \\ =x{e}^x-e^x+e^x \\ =x{e}^x \\ \text{Given:}x>0 \\ \mathrm{W}\text{e know,} \\ {e}^x>0 \\ \text{⇒}x{e}^x>0 \\ \Rightarrow f\text{'}\left(x\right)>0,\forall x>0 \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is increasing on for all}\mathit{\text{x}}\text{>0.} \end{gathered}$$