just differentiate and solve. What's the trouble?
f = 3x^3*e^-x
f' = (9x^2 - 3x^3)e^-x
= 3x^2(3-x)e^-x
Now, x^2 >= 0 and e^-x >= 0
so the only way we can have f' < 0 is x>3
That means choice (e)
Check: 3*4.797^2*(3-4.797)e^-4.797 = -1.024
let f be the function given by f(x)=(3x^3/e^x). for what value of x is the slope of the line tangent to f equal to -1.024?
a)-9.004
b)-4.734
c)1.029
d)1.277
e)4.797
1 answer
1 answer