Find maximum and minimmn values of f(x)=1-e⁻ˣ x ≥ 0
Answers
Answered by
0
any function, f(x) attains maximum at x = a only when f'(a) = 0 and f"(a) < 0 while function , f(x) attains minimum at x = b only when f'(b) = 0 and f"(b) > 0.
f(x) = 1 - e^-x , x ≥ 0
differentiate f(x) with respect to x,
f'(x) = 0 - (-1)e^-x = e^-x
now, f'(x) = 0
e^-x ≠ 0 , exponential function doesn't equal to zero.
hence, given function is neither maximum nor minimum.
f(x) = 1 - e^-x , x ≥ 0
differentiate f(x) with respect to x,
f'(x) = 0 - (-1)e^-x = e^-x
now, f'(x) = 0
e^-x ≠ 0 , exponential function doesn't equal to zero.
hence, given function is neither maximum nor minimum.
Attachments:
Answered by
0
Dear Student:
F(x)=f(x)=1-e⁻ˣ x ≥ 0
For,maximum and minimum
We will find df/dx and then equate to zero.
And then again proceed for second derivative and see it is negative or positive.
If it is positive then f will be minimum
And if negative then f maximum
See the attachment:
Attachments:
Similar questions