Find the maximum and minimum values of the fu | vedclass

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(B) The given function is $f(x)=(2x-1)^{2}+3$.It is known that for any real number $x$,the square of a real number is always non-negative,i.e.,$(2x-1)

Vedclass · Accepted Answer

Minimum value is $3$,maximum value does not exist.

Vedclass · Answer

(B) The given function is $f(x)=(2x-1)^{2}+3$.It is known that for any real number $x$,the square of a real number is always non-negative,i.e.,$(2x-1)^{2} \geq 0$.Adding $3$ to both sides,we get $(2x-1)^{2}+3 \geq 0+3$,which means $f(x) \geq 3$ for all $x \in \mathbb{R}$.The minimum value of $f(x)$ is attained when $(2x-1)^{2} = 0$.Setting $2x-1 = 0$,we get $x = \frac{1}{2}$.Thus,the minimum value is $f\left(\frac{1}{2}ight) = (2 \cdot \frac{1}{2} - 1)^{2} + 3 = 0 + 3 = 3$.Since the function $(2x-1)^{2}$ increases indefinitely as $x 	o \infty$ or $x 	o -\infty$,the function $f(x)$ does not have a maximum value.

Find the maximum and minimum values of the function $f(x)=(2x-1)^{2}+3$.

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