Question: A Random Variable that is normally distributed with mean $\mu$ and standard deviation $\sigma$ has a probability density function of
<p>$$ f(x) = {1\over {\sigma}{\sqrt {2\pi}}} e^{(x-\mu)^2\over 2 \sigma^2} $$</p> <p>Find the values of $x$ where the curve has points of inflection </p>
What I have attempted:
$$ f(x) = {1\over {\sigma}{\sqrt {2\pi}}}e^{(x-\mu)^2\over 2{\sigma}^2} $$
$$ f'(x) = {1\over \sigma{\sqrt {2\pi}}}e^{(x-\mu)^2\over 2{\sigma}^2} \cdot {(x-\mu)\over \sigma^2} $$
$$ f''(x) = \left[({1\over \sigma{\sqrt {2\pi}}}e^{(x-\mu)^2\over 2\sigma^2} \cdot {(x-\mu)\over \sigma^2} \cdot {(x-\mu)\over \sigma^2}) + {1\over \sigma{\sqrt {2\pi}}}e^{(x-\mu)^2\over 2\sigma^2} \cdot {1\over \sigma^2}\right] = 0 $$
$$ ({1\over {\sigma}{\sqrt {2\pi}}}e^{(x-\mu)^2\over 2{\sigma}^2}) \left[ {(x-\mu)^2\over {\sigma}^4} + {1\over \sigma^2} \right] = 0 $$
$$ \left[ {(x-\mu)^2\over {\sigma}^4} + {1\over \sigma^2} \right] = 0 $$
$$ \left[ {(x-\mu)^2\over {\sigma}^4} + {\sigma^2\over \sigma^4} \right] = 0 $$
$$ {(x-\mu)^2} + {\sigma^2\ } = 0 $$
$$ x^2 - 2{\mu}x + \mu^2 + {\sigma^2\ } = 0 $$
$$ x={-b\pm\sqrt{b^2-4ac} \over 2a} $$
$$ x={2\mu\pm\sqrt{4\mu^2-4(\mu^2 + \sigma^2) } \over 2} $$
$$ x={2\mu\pm\sqrt{4\mu^2-4\mu^2 - 4 \sigma^2 } \over 2} $$
$$ x={2\mu\pm\sqrt{ - 4 \sigma^2 } \over 2} $$
$$ x={2\mu\pm\ 2\sigma i \over 2} $$
$$ x = \mu ± \sigma i $$
Is this correct? I'm doubting myself due to the fact I have an imaginary number...
