If f(x) = begin{cases} -x^3 + 1, & text{if } | vedclass

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(D) Given the function $f(x) = \begin{cases} -x^3 + 1, & x \leq 1 \ |x - 1| + \lambda, & x > 1 \end{cases}$.At $x = 1$,the value of the function is $

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$f(x)$ has a point of minima at $x = 1, \forall \lambda > 0$

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(D) Given the function $f(x) = \begin{cases} -x^3 + 1, & x \leq 1 \ |x - 1| + \lambda, & x > 1 \end{cases}$.At $x = 1$,the value of the function is $f(1) = -(1)^3 + 1 = 0$.For $x > 1$,$f(x) = |x - 1| + \lambda = (x - 1) + \lambda$.As $x 	o 1^+$,$f(x) 	o \lambda$.For $x = 1$ to be a point of minima,we must have $f(1) \leq f(x)$ for all $x$ in a neighborhood of $1$.Since $f(1) = 0$,we require $0 \leq \lambda$ for the function to be continuous or to maintain the minimum at $x=1$ relative to the right-hand side limit.Specifically,if $\lambda > 0$,then for $x$ slightly greater than $1$,$f(x) = (x-1) + \lambda > 0 = f(1)$.Thus,$f(x)$ has a point of minima at $x = 1$ for all $\lambda > 0$.

If $f(x) = \begin{cases} -x^3 + 1, & \text{if } -\infty < x \leq 1 \\ |x - 1| + \lambda, & \text{if } x > 1 \end{cases}$,then:

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