Find the stationary point of y=f(x) when f(x)=3e^(-x^2) and hence show that d^2/dy^2= 6e^(-x^2) x (2x-1)
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You need to evaluate the first order derivative, using the chain rule, such that:
`f'(x) = (3e^(-x^2))*(-x^2)' => f'(x) = -6x*e^(-x^2)`
Notice that the first derivative is negative for x>0 and it is positive for x < 0.
You need to determine the second derivative using the product rule, such that:
`f''(x) = (-6x * e^(-x^2))''`
`f''(x) = (-6x)' * (e^(-x^2)) + (-6x) * (e^(-x^2))'`
`f''(x) = -6 * (e^(-x^2)) - 6x * e^(-x^2) * (-2x)`
`f''(x) = -6 * (e^(-x^2)) + 12x^2 * e^(-x^2)`
You need to factor out `6e^(-x^2)` such that:
`f''(x) =6e^(-x^2) * (-1 + 2x^2)`
Hence, evaluating the expression of the second derivative of the function yields `f''(x) =6e^(-x^2) * (2x^2 - 1)` , but not `f''(x) =6e^(-x^2) * (2x - 1).`
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