Answer:
(C) f’(c) = 0 and f”(c) > 0
Step-by-step explanation:
A minimum occurs where the first derivative is 0 (the tangent line is horizontal), and the second derivative is positive (concave up). The simplest example of this is a positive parabola, like y = x², which has a relative minimum at its vertex.
Answer:
it does have a solution x=0
Step-by-step explanation:
9x+15=3x+15
6x+15=15
6x=0
x=0
The area between the two functions is 0
<h3>How to determine the area?</h3>
The functions are given as:
f₁(x)= 1
f₂(x) = |x - 2|
x ∈ [0, 4]
The area between the functions is
A = ∫[f₂(x) - f₁(x) ] dx
The above integral becomes
A = ∫|x - 2| - 1 dx (0 to 4)
When the above is integrated, we have:
A = [(|x - 2|(x - 2))/2 - x] (0 to 4)
Expand the above integral
A = [(|4 - 2|(4 - 2))/2 - 4] - [(|0 - 2|(0 - 2))/2 - 0]
This gives
A = [2 - 4] - [-2- 0]
Evaluate the expression
A = 0
Hence, the area between the two functions is 0
Read more about areas at:
brainly.com/question/14115342
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