(i) f(x) is continuous and defined for all re | vedclass

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(B) From the given information:$1$. $f'(-5) = 0$ and $f'(4) = 0$ implies stationary points at $x = -5$ and $x = 4$.$2$. $f'(x) > 0$ for $x < -5$ and $

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(B) From the given information:$1$. $f'(-5) = 0$ and $f'(4) = 0$ implies stationary points at $x = -5$ and $x = 4$.$2$. $f'(x) > 0$ for $x $3$. $f'(2)$ is undefined,indicating a sharp corner or cusp at $x = 2$.$4$. $f'(x) > 0$ for $2  4$ implies a local maximum at $x = 4$.$5$. $f''(x) Comparing these properties with the given options:- Option $A$ shows a local minimum at $x = -5$,which is incorrect.- Option $B$ shows a local maximum at $x = -5$ and a local maximum at $x = 4$,with a cusp at $x = 2$,and the graph is concave down throughout. This matches all conditions.- Option $C$ shows a local maximum at $x = -5$ but a local minimum at $x = 4$,which is incorrect.- Option $D$ shows a local maximum at $x = -5$ but a local minimum at $x = 4$,which is incorrect.Thus,the correct graph is $B$.

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