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

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(D) Given $f(x) = x^3 + 1$.Step $1$: Find the derivative $f'(x) = 3x^2$.Step $2$: For a local extremum,we set $f'(x) = 0$,which gives $3x^2 = 0$,so $x

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

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(D) Given $f(x) = x^3 + 1$.Step $1$: Find the derivative $f'(x) = 3x^2$.Step $2$: For a local extremum,we set $f'(x) = 0$,which gives $3x^2 = 0$,so $x = 0$.Step $3$: 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 at $x = 0$,the function $f(x)$ does not have a local extremum at $x = 0$. Thus,Statement-$1$ is false.Step $4$: The function $f(x) = x^3 + 1$ is a polynomial function,which is continuous and differentiable on $(-\infty, \infty)$. Also,$f'(0) = 3(0)^2 = 0$. Thus,Statement-$2$ is true.

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