Find the intervals from which f is increasing or decreasing for f(x)=3x^2/5-x^3/5
1 answer:
We get the derivative :
f'(x) is negative for x<0.
Let's look at the critical points other than at x=0:
f'(x)=0 <=> 6/5-(3/5)x^(1/5)=0 <=> x^(1/5)=(6/5)*(5/3)=2 hence x=2^5=32
There is a critical point at x=32; for x<=32 f' is positive, for x>=30 f' is negative.
The derivative is positive for positive xs and negative for negative xs, hence f is increasing for 32>=x>0, and decreasing for x<0 and x>32
You might be interested in
Let's start by tidying up that equation and put it into slope-intercept form (y = mx + b); from there, we can plug in coordinates.
Let's use the distributive property on the right side:
Now add 4 to both sides
Which simplifies to:
Since that's the equation of our line, now we can plug in coordinates and see what it churns out.
We know that the x-coordinate of P = 4 so let's substitute 4 in for x and calculate the y-coordinate:
So the y-coordinate for point P = 10
Let's use the distributive property on the right side:
Now add 4 to both sides
Which simplifies to:
Since that's the equation of our line, now we can plug in coordinates and see what it churns out.
We know that the x-coordinate of P = 4 so let's substitute 4 in for x and calculate the y-coordinate:
So the y-coordinate for point P = 10