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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.

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 $f'(0) = 0$.