Find the maximum and minimum values of the fu | vedclass

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(A) The given function is $f(x) = -(x-1)^{2} + 10$. We know that for any real number $x$,$(x-1)^{2} \geq 0$. Multiplying by $-1$,we get $-(x-1)^{2} \l

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Maximum value is $10$,minimum value does not exist.

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(A) The given function is $f(x) = -(x-1)^{2} + 10$. We know that for any real number $x$,$(x-1)^{2} \geq 0$. Multiplying by $-1$,we get $-(x-1)^{2} \leq 0$. Adding $10$ to both sides,we get $-(x-1)^{2} + 10 \leq 10$. Thus,$f(x) \leq 10$ for all $x \in R$. The maximum value is attained when $(x-1)^{2} = 0$,which implies $x = 1$. Therefore,the maximum value is $f(1) = -(1-1)^{2} + 10 = 10$. Since the function $f(x) = -(x-1)^{2} + 10$ is a downward-opening parabola,as $x 	o \infty$ or $x 	o -\infty$,$f(x) 	o -\infty$. Hence,the function does not have a minimum value.

Find the maximum and minimum values of the function given by $f(x) = -(x-1)^{2} + 10$.

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