find the intervals for which the function f(x)=x(x-1)^3 is strictly increasing or decreasing ...step by step explanation....PLSS DON'T SPAM..IF U DON'T KNOW THE ANSWER THEN DON'T TAKE THE CHANCES OF OTHERS WHO KNOW THE ANSWER TO THIS QUESTION PLSSS
Answers
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Answer:
Step-by-step explanation:
Let's find the intervals where f(x)=x^3+3x^2-9x+7f(x)=x
3
+3x
2
−9x+7f, left parenthesis, x, right parenthesis, equals, x, cubed, plus, 3, x, squared, minus, 9, x, plus, 7 is increasing or decreasing. First, we differentiate fff:
f'(x)=3x^2+6x-9f
′
(x)=3x
2
+6x−9f, prime, left parenthesis, x, right parenthesis, equals, 3, x, squared, plus, 6, x, minus, 9 [Show entire calculation]
Now we want to find the intervals where f'f
′
f, prime is positive or negative.
f'(x)=3(x+3)(x-1)f
′
(x)=3(x+3)(x−1)f, prime, left parenthesis, x, right parenthesis, equals, 3, left parenthesis, x, plus, 3, right parenthesis, left parenthesis, x, minus, 1, right parenthesis
f'f
′
f, prime intersects the xxx-axis when x=-3x=−3x, equals, minus, 3 and x=1x=1x, equals, 1, so its sign must be constant in each of the following intervals:
Let's evaluate f'f
′
f, prime at each interval to see if it's positive or negative on that interval.
Interval xxx-value f'(x)f
′
(x)f, prime, left parenthesis, x, right parenthesis Verdict
x<-3x<−3x, is less than, minus, 3 x=-4x=−4x, equals, minus, 4 f'(-4)=15>0f
′
(−4)=15>0f, prime, left parenthesis, minus, 4, right parenthesis, equals, 15, is greater than, 0 fff is increasing. \nearrow↗\nearrow
-3<x<1−3<x<1minus, 3, is less than, x, is less than, 1 x=0x=0x, equals, 0 f'(0)=-9<0f
′
(0)=−9<0f, prime, left parenthesis, 0, right parenthesis, equals, minus, 9, is less than, 0 fff is decreasing. \searrow↘\searrow
x>1x>1x, is greater than, 1 x=2x=2x, equals, 2 f'(2)=15>0f
′
(2)=15>0f, prime, left parenthesis, 2, right parenthesis, equals, 15, is greater than, 0 fff is increasing. \nearrow↗\nearrow
So fff is increasing when x<-3x<−3x, is less than, minus, 3 or when x>1x>1x, is greater than, 1 and increasing when -3<x<1−3<x<1minus, 3, is less than, x, is less than, 1.