Let fx - 1 - sin x- x 0 x2 - x - 1- x - 0- Then

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First Derivative Test for Local Maximum

Let fx = 1 + ...

Question

Let f(x) = ${\begin{matrix} 1 + s i n x, x < 0 x^{2} - x + 1, x \geq 0 \end{matrix}$ . Then

f has a local maximum at x = 0

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f has a local minimum at x = 0

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f is increasing every where

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f is decreasing everywhere

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Solution

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Similar questions

f (x) = {\begin{matrix} \begin{matrix} 0, & x < 0 \end{matrix} \begin{matrix} x^{2}, & x \geq 0 \end{matrix} \end{matrix}

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x = 0

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{\begin{matrix} (x - 1)^{1 / 3}, x < 0 (x + 1)^{1 / 2}, x \geq 0 \end{matrix}

f (x) = {\begin{matrix} - 1, - 2 \leq x < 0 x^{2} - 1, 0 \leq x \leq 2 \end{matrix}

g (x) = | f (x) | + f (| x |)

(- 2, 2), g

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= {\begin{matrix} 2 x + 1; x \leq 0 x^{2}; x > 0 \end{matrix}

f (- 1), f (1)

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