Aspire Faculty ID #16721 · Topic: CUET 2023 · Just now
CUET 2023

If $f:\mathbb{R}\to\mathbb{R}$ is defined as $f(x)=x^2+1$, then the minimum value of $f(x)$ is

Solution

Since $x^2\ge0$ for all real $x$, $f(x)=x^2+1\ge1$ Minimum occurs at $x=0$.

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