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Q. If $y = x^3 + x^2 + x + 1$, then $y$

Application of Derivatives

Solution:

Let $f(x) = y = x^3 + x^2 + x+ 1$. Then,
$f'(x ) = 3x^2 + 2 x + 1$.
For a maximum or minimum, we have
$f'(x) = 0 \Rightarrow 3x^2 + 2x+ 1 = 0$
But, this equation gives imaginary values of $x$.
So, $f'(x) \ne 0$ for any real value of $x$.
Hence, $f(x)$ does not have a maximum or minimum.