The values of x at the stationary points of f | vedclass

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(C) Given the function $f(x) = x^3 + 3x^2 - 2$. To find the stationary points,we calculate the first derivative of the function with respect to $x$: $

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$0, -2$

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(C) Given the function $f(x) = x^3 + 3x^2 - 2$. To find the stationary points,we calculate the first derivative of the function with respect to $x$: $f'(x) = \frac{d}{dx}(x^3 + 3x^2 - 2) = 3x^2 + 6x$. Stationary points occur where the first derivative is equal to zero: $f'(x) = 0 \Rightarrow 3x^2 + 6x = 0$. Factoring the expression,we get: $3x(x + 2) = 0$. This gives the values $x = 0$ and $x = -2$. Therefore,the values of $x$ at the stationary points are $0$ and $-2$.

The values of $x$ at the stationary points of $f(x)=x^3+3x^2-2$ are

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