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

For all $x \in \mathbb{R}$, the minimum value $\frac{1}{3}$ and the maximum value $3$ of $f(x) = \frac{x^2+x+1}{x^2-x+1}$ occur at $l$ and $m$ respectively. Then $l+m$ is equal to:

If the area of a right-angled triangle with hypotenuse $5$ is maximum,then its perimeter is

If $A=\{x : 9x \geq x^2+20\}$ and $f: A \rightarrow R$ is defined by $f(x)=2x^3-15x^2+36x-48$,then the maximum value of $f(x)$ is

Show that the semi-vertical angle of a right circular cone of given surface area and maximum volume is $\sin ^{-1}\left(\frac{1}{3}\right)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module