If A={x : 9x geq x^2+20} and f: A rightarrow | vedclass

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

Question

(B) Given $A=\{x : 9x \geq x^2+20\}$.Solving the inequality $x^2-9x+20 \leq 0$,we get $(x-4)(x-5) \leq 0$,which implies $x \in [4, 5]$.Thus,$A=[4, 5]$

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) Given $A=\{x : 9x \geq x^2+20\}$.Solving the inequality $x^2-9x+20 \leq 0$,we get $(x-4)(x-5) \leq 0$,which implies $x \in [4, 5]$.Thus,$A=[4, 5]$.Given $f(x)=2x^3-15x^2+36x-48$.Finding the derivative,$f'(x)=6x^2-30x+36=6(x^2-5x+6)=6(x-3)(x-2)$.For critical points,set $f'(x)=0$,which gives $x=2$ and $x=3$.Since both critical points $x=2$ and $x=3$ lie outside the interval $A=[4, 5]$,the function $f(x)$ is monotonic in this interval.Checking the sign of $f'(x)$ for $x \in [4, 5]$: $f'(4)=6(4-3)(4-2)=12 > 0$.Since $f'(x) > 0$ for all $x \in [4, 5]$,$f(x)$ is an increasing function on $[4, 5]$.Therefore,the maximum value occurs at the right endpoint $x=5$.$f(5)=2(5)^3-15(5)^2+36(5)-48 = 2(125)-15(25)+180-48 = 250-375+180-48 = 7$.

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

Similar Questions

Local maximum and local minimum values of the function $f(x) = (x - 1)(x + 2)^2$ are

The difference between the greatest and the least values of the function $f(x) = x(\ln x - 2)$ on the interval $[1, e^2]$ is:

$y = 2x^3 - 8x^2 + 10x - 4$ is a function defined on $[1, 2]$. If the tangent drawn at a point $(a, b)$ on the graph of this function is parallel to the $X$-axis and $a \in (1, 2)$, then $a =$

Find all the points of local maxima and local minima of the function $f$ given by $f(x) = 2x^3 - 6x^2 + 6x + 5$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module