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What is the least value of the function $f(x) = 2x^{2}-8x-3$ in the interval $[0, 5]$?

1. $-15$
2. $7$
3. $-11$
4. $-3$
in Calculus | 2k views

$f(x) = 2x^{2}-8x-3$

$f'(x) = 4x-8$

For stationary point$:f'(x) = 0\implies 4x - 8 = 0 \implies x = 2$

Therefore, critical points $x = 0,2,5$

Now, $f''(x) = 4>0\:\: \textbf{(Minima)}$

For getting the minimum (or) least value , we should check all the value of critical points (stationary point and closed interval points).

For $x = 0: f(x) = -3$

For $x = 2: f(x) = 8 - 16 - 3 = -11$

For $x = 5: f(x) = 50 - 40 - 3 = 7$

$\therefore x= 2 ,\text{minimum value of}\: f(x) = -11$

So, the correct answer is $(C).$
by Veteran (58.9k points)
selected by

f(x) = 2x2 -8x -3

f'(x)=4x-8

f''(x)=4 , then it has minimal value at the point x=2

minimum value is 2⨉(2)2 -8⨉(2) -3= -11

by Veteran (119k points)
0
YOU have forgot to check the minimum for end points
0
not getting @viv

Substutue all the value of x from 1 to 5

in case of 0 it is -3

in case of 1 you will get -9

in case of 2 you will get -11

in case of 3 you will get -9

incase of 4 you will get -3

incase of 5 you will get 7

so the least value is -11 -->option c

by Loyal (9.9k points)
0
how u have taken the value of x from 1 to 5
0
Unreliable way
+2
reliable but brute force, in ISRO time will play an important role and you can solve some other easy questions then solving this way. each question carry same mark!