Find the extreme values of the function and where they occur.
-$$y = \frac { 1 } { x ^ { 2 } + 1 }$$
A) The maximum value is 1 at x = 0.5, the minimum value is -1 at x = 0.5.
B) The minimum value is -1 at x = 0.5.
C) The maximum value is 1 at x = 0.5.
D) The maximum value is 1 at x = 0.

Question

Quizplus · Accepted Answer

The function $y = \frac{1}{x^2 + 1}$ has a maximum value when the denominator is at its minimum. Since $x^2$ is always non-negative, the smallest value $x^2 + 1$ can have is 1, which occurs when $x = 0$. Thus, the maximum value of $y$ is $\frac{1}{1} = 1$, occurring at $x = 0$. There is no minimum value other than the limit as $x$ approaches infinity, which tends towards 0 but never reaches a specific minimum value at a finite $x$.

Find the Extreme Values of the Function and Where They y=1x2+1y = \frac { 1 } { x ^ { 2 } + 1 }y=x2+11​

Find the Extreme Values of the Function and Where They $y = \frac { 1 } { x ^ { 2 } + 1 }$