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) }=0 ... eqn (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