The minimum value of f(x) = e^{(x^4 - x^3 + x | vedclass

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(C) Given function is $f(x) = e^{(x^4 - x^3 + x^2)}$.To find the minimum value,we find the derivative $f'(x)$:$f'(x) = e^{(x^4 - x^3 + x^2)} \cdot \fr

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$1$

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(C) Given function is $f(x) = e^{(x^4 - x^3 + x^2)}$.To find the minimum value,we find the derivative $f'(x)$:$f'(x) = e^{(x^4 - x^3 + x^2)} \cdot \frac{d}{dx}(x^4 - x^3 + x^2)$$f'(x) = e^{(x^4 - x^3 + x^2)} \cdot (4x^3 - 3x^2 + 2x)$$f'(x) = e^{(x^4 - x^3 + x^2)} \cdot x(4x^2 - 3x + 2)$.For the quadratic factor $4x^2 - 3x + 2$,the discriminant $D = (-3)^2 - 4(4)(2) = 9 - 32 = -23 Since the leading coefficient $4 > 0$ and $D Thus,the sign of $f'(x)$ depends only on $x$.For $x For $x > 0$,$f'(x) > 0$,so the function is increasing.Therefore,the function attains its minimum value at $x = 0$.The minimum value is $f(0) = e^{(0^4 - 0^3 + 0^2)} = e^0 = 1$.

The minimum value of $f(x) = e^{(x^4 - x^3 + x^2)}$ is

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