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2 years ago

What is the maximum slope to f(x)= e^(-x^2)?

1 answer:

6 0

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

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1 year ago

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