f(x)=-8xlnx decreases over what domain
f'(x) = -8x(1/x ) + (-8)lnx
= -8 - lnx
to be decreasing, -8 - 8lnx < 0
8lnx + 8 > 0
8 lnx > -8
lnx > -1
(solve lnx = -1, x = 1/e
which is true for all x > 1/e
To find the domain over which the function f(x) = -8xln(x) decreases, we need to determine where the derivative of the function is negative.
First, let's find the derivative of f(x) with respect to x using the product rule and chain rule:
f'(x) = -8ln(x) - 8x * (1/x)
Now, let's simplify the derivative:
f'(x) = -8ln(x) - 8
To find where the function is decreasing, we need to solve the inequality f'(x) < 0:
-8ln(x) - 8 < 0
Divide by -8 to simplify:
ln(x) + 1 > 0
Subtract 1 from both sides:
ln(x) > -1
To eliminate the logarithm, we use exponentiation with a base of e:
x > e^(-1)
Therefore, the domain over which f(x) = -8xln(x) decreases is x > e^(-1), or x > 1/e.