Find the maximum and minimum values of the fu | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $f(x)=(2x-1)^{2}+3$.

Similar Questions

At what value of $x$ does the function $f(x) = \sin x(1 + \cos x)$ have a maximum value?

The function $f(x) = \sin x (1 + \cos x)$ is maximum at $x = \dots$

$A$ wire of length $28 \, m$ is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

Find the absolute maximum and minimum values of the function $f$ given by $f(x) = \cos^{2} x + \sin x$,where $x \in [0, \pi]$.

Find the maximum and minimum values of the function given by $h(x) = x + 1, x \in (-1, 1)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module