What is the maximum slope to f(x)= e^(-x^2)? (what's max. slope? ) a. (sqrt2)/2 b. -(sqrt2)/2 c. -sqrt(2/e) d. sqrt(2/e)?

F'(x)=-2x{e^(-x²)}
Equating f'(x)=0,
i.e,

-2x{e^(-x²)}=0eqn(i)
For maximum slope:
Differentiating (i) wrt x,we get,
4x² {e^(-x²)}-2 {e^(-x²)}=0
{e^(-x²)} [4x²-2]=0
4x²-2=0
x=1/√2

If f(x)=y then f'(x)=dy/dx=slope
So,
Maximum slope
= f'(x)
= -2x{e^(-x²)}
=-2×(1/✓2){e^(-1/✓2)²}
= -✓2{e^(-1/2)}
=(-sqrt 2/sqrt e)
=-sqrt(2/e)➡c is correct

29. because you will divide each number of months by 12 to find the number of years. which would mean, 5, 12.5, 16, 45, 50, 72.5. which is crazy. anyhow, i think you find the median in the middle. and yea hope your test goes good