Let f: ( -infty, infty ) to ( -infty, infty ) | vedclass

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(D) Given the function $f(x) = x^3 + 1$.To check for local extrema,we find the derivative: $f'(x) = 3x^2$.Setting $f'(x) = 0$,we get $3x^2 = 0$,which

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Statement $1$ is false,Statement $2$ is true.

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(D) Given the function $f(x) = x^3 + 1$.To check for local extrema,we find the derivative: $f'(x) = 3x^2$.Setting $f'(x) = 0$,we get $3x^2 = 0$,which implies $x = 0$.Now,we check the sign of $f'(x)$ around $x = 0$:For $x  0$.For $x > 0$,$f'(x) = 3x^2 > 0$.Since the derivative $f'(x)$ does not change sign as $x$ passes through $0$,the function $f(x)$ does not have a local extremum at $x = 0$. Thus,Statement $1$ is false.For Statement $2$: The function $f(x) = x^3 + 1$ is a polynomial function,which is continuous and differentiable everywhere on $( -\infty, \infty )$.Also,$f'(x) = 3x^2$,so $f'(0) = 3(0)^2 = 0$. Thus,Statement $2$ is true.Therefore,Statement $1$ is false and Statement $2$ is true.

List-$I$	List-$II$
$A. \frac{d}{dx}\left(\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\right)$	$(i) \log(x+\sqrt{1+x^2})$
$B. \frac{d}{dx}\left(\frac{3+\|x-1\|}{3x+4}\right)$	$(ii) -\frac{4x}{(1+x^2)^2}$
$C. \sinh^{-1} x$	$(iii) \frac{1}{2}$
$D. \frac{d^2}{dx^2}\left(\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right)$	$(iv) \frac{1}{\sqrt{1+x^2}}$
	$(v) \text{not differentiable at } x=1$

Let $f: ( -\infty, \infty ) \to ( -\infty, \infty )$ be defined by $f(x) = x^3 + 1$.
Statement $1$: The function $f$ has a local extremum at $x = 0$.
Statement $2$: The function $f$ is continuous and differentiable on $( -\infty, \infty )$ and $f'(0) = 0$.

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Let $f: ( -\infty, \infty ) \to ( -\infty, \infty )$ be defined by $f(x) = x^3 + 1$.Statement $1$: The function $f$ has a local extremum at $x = 0$.Statement $2$: The function $f$ is continuous and differentiable on $( -\infty, \infty )$ and $f'(0) = 0$.