Respuesta :
Answer:
a) decreases at interval (-∞,5) and increases at (5,∞)
b) is convave down at interval (-∞,6) an up at interval (6,∞)
c) f(x) has an inflexion point at x=6
Step-by-step explanation:
a) for the function
f(x) = (4 − x)*e^(−x)
then the derivative of f(x) indicates if the function decreases or increases. Thus
f'(x) =df(x)/dx = -e^(−x) -(4 − x)*e^(−x)= (x-5)*e^(−x)
since e^(−x) is always positive , then
f'(x) < 0 for x<5 → f(x) decreases when x<5 ( interval (-∞,5) )
f'(x) > 0 for x>5 → f(x) increases when x>5 ( interval (5,∞) )
f'(x) = 0 for x=5 → f(x) has a local minimum ( since first decreases and then increases)
b) the concavity is found with the second derivative of f(x) , then
f''(x) =d²f(x)/(dx)² = e^(−x) - (x-5)*e^(−x) = (6-x)*e^(−x)
then
f''(x) < 0 for x>6 → f(x) is convave up for x>6 ( interval (6,∞) )
f'(x) > 0 for x<6 → f(x) is concave down when x<6 ( interval (-∞,6) )
f'(x) = 0 for x=6 → f(x) has an inflection point at x=6
The function is decreases at interval (-∞,5) and increases at (5,∞),
The function is concave down at interval (-∞,6) an up at interval (6,∞),
f(x) has an inflexion point at x=6 for the function.
Given
Function = f(x) = (4 − x).[tex]e^{-x}[/tex]
According to the question;
- The derivative of f(x) indicates if the function decreases or increases. Thus
f'(x) =[tex]\frac{df (x)}{dx} = e^{-x} - (4-x). (e^{-x}) = (x-5). e^{-x}[/tex]
Where, [tex]e^{-x}[/tex] is always positive ,
Then
f'(x) < 0 for x<5 → f(x) decreases when x<5 interval (-∞,5)
f'(x) > 0 for x>5 → f(x) increases when x>5 interval (5,∞)
f'(x) = 0 for x=5 → f(x) has a local minimum ( since first decreases and then increases)
- The concavity is found with the second derivative of f(x) ,
Then,
f''(x) = = - (x-5)* = (6-x).
f''(x) < 0 for x>6 → f(x) is concave up for x>6 interval (6,∞)
f'(x) > 0 for x<6 → f(x) is concave down when x<6 interval (-∞,6)
- f'(x) = 0 for x=6 → f(x) has an inflection point at x=6.
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