# How do you find all points of inflection given y=-x^4+3x^2-4?

Nov 25, 2016

$\left(1 , - 2\right) \mathmr{and} \left(- 1 , - 2\right) \text{ }$are two inflection points of the given function.

#### Explanation:

The inflection points of a function is determined by computing
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the second derivative of $y$ then solving for $\textcolor{b l u e}{y ' ' = 0}$.
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$y = - {x}^{4} + 3 {x}^{2} - 4$
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$y ' = - 4 {x}^{3} + 6 x$
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$y ' ' = - 12 {x}^{2} + 6$
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To find the inflection points we would solve the equation:
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$\textcolor{b l u e}{y ' ' = 0}$
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$\Rightarrow - 12 {x}^{2} + 6 = 0$
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$\Rightarrow - 12 \left({x}^{2} - 1\right) = 0$
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$\Rightarrow - 12 \left(x - 1\right) \left(x + 1\right) = 0$
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$\Rightarrow x - 1 = 0 \Rightarrow x = 1 \text{ }$
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Or
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$x + 1 = 0 \Rightarrow x = - 1$
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The ordinate of the point of abscissa $x = 1$ is:
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${y}_{\left(x = 1\right)} = - {1}^{4} + 3 {\left(1\right)}^{2} - 4 = - 1 + 3 - 4 = - 2$
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The ordinate of the point of abscissa $x = - 1$ is:
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${y}_{\left(x = - 1\right)} = - {\left(- 1\right)}^{4} + 3 {\left(- 1\right)}^{2} - 4 = - 1 + 3 - 4 = - 2$
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Hence,
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$\left(1 , - 2\right) \mathmr{and} \left(- 1 , - 2\right) \text{ }$ are two inflection points of the given function.