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

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 $f(x) = -(x-1)^{2} + 10$.

Similar Questions

If a cylindrical vessel of given volume $V$ with no lid on the top is to be made from a sheet of metal,then the radius $(r)$ and height $(h)$ of the vessel so that the metal sheet used is minimum,is

Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius $r$ is $\frac{4r}{3}$.

An enemy fighter jet is flying along the curve given by $y = x^2 + 2$. $A$ soldier is placed at $(3, 2)$ and wants to shoot down the jet when it is nearest to him. The nearest distance is:

Let $f(x) = \int_{0}^{x} e^{x+t} dt$. Then the abscissa of the point where the tangent to $f(x)$ is parallel to the $x$-axis is:

$A$ wire $40 \text{ m}$ in length is to be cut into two pieces. One piece is formed into a square and the other piece into a circle. What should be the lengths of the two pieces so that the combined area of the square and the circle is minimum?

Vedclass Test Series

Exam Paper Generator

Online Exam Module