Find the relative extreme points of the tunction, it they exist Then skelch a gr
ID: 3145021 • Letter: F
Question
Explanation / Answer
Q.1 F(x) = (x-3)1/3
f'(X) = 1/3 (x-3)-2/3
f'(X) = 0 for relative minimum points
= > 1/3(x-3)-2/3 =0
so for no value of x the f'(X) where f'(x) is zero here so there are no mimum relative points.
Graph B is correct here out of 4 option.
Question 2.
f(x) = 4x + 2 ; x E [-4.5]
Here the abolute minimum value is (-4 * 4 +2 = -14) and absolute maximum value is (4 * 5 + 2 = 22)
so option A is correct.
Question 3 .
G(X) = -x3 + 2x2 + 15x + 3
G'(x) = -3x2 + 4x + 15
so G'(x) = 0
-3x2 + 4x + 15 = 0
x = -5/3, 3
f''(X) = -6x + 4 so
f''(-5/3) = 14; f'(3) = -14
so one relative minimum point that is x = -5/3 and the f(X)minimum = -11.8148
One relative mimimum point that is x = 3 and f(x)maximum = 39
Graph C is correct.
Question 4
C(X) = 4800 + 200X
R(X) = -x2 /2 + 400x
P(X) = R(X) - C(X) = -x2 /2 + 200x - 4800
Profit will be maximum when P'(x) = 0 and P''(x) =-ve
soP'(x) = -x + 200 and P''(X) = -1 (-ve)
so at x = 200 it will be maximum
P(200) = -(200)2 /2 + 200 * 200 - 4800 = $ 15200 (answer of part C)
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