Question

Find the range of the function defined by
$f\left(x\right)=\{\begin{array}{cc}\frac{3}{3^{x^2+1}},& ifx\ge 0\\ 5x+3,& ifx<0\end{array}$

• A. $y\le 0$
• B. $0<y<3$
• C. $y<3$
• D. $3\le y\le 5$
• E. All real numbers

Answer 1

The correct option is C $y<3$

• For $x\ge 0$ , $x^2\ge o$ , $x^2+1\ge 1$ , which implies $3^{x^2+1}\ge 3$
• Therefore $y=\frac{3}{3^{x^2+1}}\le 1$ for all $x\ge 0$
• For $x<0$ , $y=5x+3<3$
• Therefore the answer is $y<3$