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ASSIGNMENT NO 1 Solution
MTH-101
Q: If Find the intervals where the given function is
Solution:
f(x)=x^{3}-27x
After Taking Derivative
f `(x)=3x^{2}-27
Function will increase if it is greater than zero
f `(x)=3x^{2}-27>0
3x^{2}-27>0
x^{2}-9>0
x^{2}>9
After taking square root
x>3
So, function increase on (3,∞)
f(x)=x^{3}-27x
Take Derivative
f `(x)=3x^{2}-27
Function will decrease when it is less than zero
f `(x)=3x^{2}-27<0
3x^{2}-27<0
x^{2}-9<0
x^{2}<9
After taking square root
x<3
So, function decrease on (- ∞, 3)
f(x)=x^{3}-27x
After Taking Derivative
f `(x)=3x^{2}-27
After taking second derivative
f ``(x)=6x
f ``(x)=6x>0
6x>0
x>0
So, it is concave up on (0,+∞)
f(x)=x^{3}-27x
After Taking Derivative
f `(x)=3x^{2}-27
After taking second derivative
f ``(x)=6x
f ``(x)=6x<0
6x<0
x<0
So, it is concave down on (- ∞,0)
POINT OF INFLECTION
f(x)=x^{3}-27x ------------------ (1)
After Taking Derivative
f `(x)=3x^{2}-27
After taking second derivative
f ``(x)=6x
f ``(x)= 6x =0
6x=0
x=0
Put this in (1)
f(0)=(0)^{3}- 27(0)
f(0)=0
So, inflection point is (0 , 0)
tnk u
Aftab gud keep it up & thanks for sharing
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