If f(x) = x^4 + lambda x^3 + x^2 (lambda in R | vedclass

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(B) Given $f(x) = x^4 + \lambda x^3 + x^2$.First,find the derivative: $f'(x) = 4x^3 + 3\lambda x^2 + 2x$.Since $f(x)$ has a local maximum at $x = \fra

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(B) Given $f(x) = x^4 + \lambda x^3 + x^2$.First,find the derivative: $f'(x) = 4x^3 + 3\lambda x^2 + 2x$.Since $f(x)$ has a local maximum at $x = \frac{1}{2}$,we have $f'(\frac{1}{2}) = 0$.$f'(\frac{1}{2}) = 4(\frac{1}{8}) + 3\lambda(\frac{1}{4}) + 2(\frac{1}{2}) = \frac{1}{2} + \frac{3\lambda}{4} + 1 = 0$.$\frac{3\lambda}{4} = -\frac{3}{2} \Rightarrow \lambda = -2$.Now,$f'(x) = 4x^3 - 6x^2 + 2x = 2x(2x^2 - 3x + 1) = 2x(2x - 1)(x - 1)$.The critical points are $x = 0, \frac{1}{2}, 1$.Using the first derivative test:For $x For $0  0$ (local minimum at $x=0$).For $\frac{1}{2} For $x > 1$,$f'(x) > 0$ (local minimum at $x=1$).Calculating values at critical points:$f(0) = 0^4 - 2(0)^3 + 0^2 = 0$.$f(1) = 1^4 - 2(1)^3 + 1^2 = 1 - 2 + 1 = 0$.Thus,the absolute minimum value of $f(x)$ is $0$.

If $f(x) = x^4 + \lambda x^3 + x^2$ $(\lambda \in R)$ has a local maximum at $x = \frac{1}{2}$,then the absolute minimum value of $f(x)$ is:

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