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extreme y=9sin(|x|),-2pi<= x<= 2pi

Solução

puntos extremos

y = 9 sin (∣ x ∣), - 2 π \leq x \leq 2 π

M \overset{a}{ˊ} ximo (- 2 π, 0), M \overset{ı}{ˊ} nimo (- \frac{3 π}{2}, - 9), M \overset{a}{ˊ} ximo (- \frac{π}{2}, 9), M \overset{ı}{ˊ} nimo (0, 0), M \overset{a}{ˊ} ximo (\frac{π}{2}, 9), M \overset{ı}{ˊ} nimo (\frac{3 π}{2}, - 9), M \overset{a}{ˊ} ximo (2 π, 0)

Passos da solução

Encontrar os pontos críticos:

x = - 2 π, x = - \frac{3 π}{2}, x = - \frac{π}{2}, x = 0, x = \frac{π}{2}, x = \frac{3 π}{2}, x = 2 π

Find intervals:

Decrescente

: - 2 π < x < - \frac{3 π}{2},

Crescente

: - \frac{3 π}{2} < x < - \frac{π}{2},

Decrescente

: - \frac{π}{2} < x < 0,

Crescente

: 0 < x < \frac{π}{2},

Decrescente

: \frac{π}{2} < x < \frac{3 π}{2},

Crescente

: \frac{3 π}{2} < x < 2 π

Substituir o ponto extremo

x = - 2 π

9 sin (∣ x ∣) \Rightarrow y = 0

M \overset{a}{ˊ} ximo (- 2 π, 0)

Substituir o ponto extremo

x = - \frac{3 π}{2}

9 sin (∣ x ∣) \Rightarrow y = - 9

M \overset{ı}{ˊ} nimo (- \frac{3 π}{2}, - 9)

Substituir o ponto extremo

x = - \frac{π}{2}

9 sin (∣ x ∣) \Rightarrow y = 9

M \overset{a}{ˊ} ximo (- \frac{π}{2}, 9)

Substituir o ponto extremo

x = 0

9 sin (∣ x ∣) \Rightarrow y = 0

M \overset{ı}{ˊ} nimo (0, 0)

Substituir o ponto extremo

x = \frac{π}{2}

9 sin (∣ x ∣) \Rightarrow y = 9

M \overset{a}{ˊ} ximo (\frac{π}{2}, 9)

Substituir o ponto extremo

x = \frac{3 π}{2}

9 sin (∣ x ∣) \Rightarrow y = - 9

M \overset{ı}{ˊ} nimo (\frac{3 π}{2}, - 9)

Substituir o ponto extremo

x = 2 π

9 sin (∣ x ∣) \Rightarrow y = 0

M \overset{a}{ˊ} ximo (2 π, 0)

M \overset{a}{ˊ} ximo (- 2 π, 0), M \overset{ı}{ˊ} nimo (- \frac{3 π}{2}, - 9), M \overset{a}{ˊ} ximo (- \frac{π}{2}, 9), M \overset{ı}{ˊ} nimo (0, 0), M \overset{a}{ˊ} ximo (\frac{π}{2}, 9), M \overset{ı}{ˊ} nimo (\frac{3 π}{2}, - 9), M \overset{a}{ˊ} ximo (2 π, 0)

f (x, y) = x y^{2} - x^{2} - y^{2}

y = 5 x - x^{2} + 66

x^{2} - 8 x + 22

\frac{x ^{4} + 8}{x ^{2}}

f (x) = (x - 2)^{3} + 5