How would you "remove the discontinuity" of f ? in other words, how would you define f(5) in order to make f continuous at 5?f(x
) = (x^2 − 2x − 15)/x − 5
1 answer:
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Answer:
x = 2 cm
y = 2 cm
A(max) = 4 cm²
Step-by-step explanation: See Annex
The right isosceles triangle has two 45° angles and the right angle.
tan 45° = 1 = x / 4 - y or x = 4 - y y = 4 - x
A(r) = x* y
Area of the rectangle as a function of x
A(x) = x * ( 4 - x ) A(x) = 4*x - x²
Tacking derivatives on both sides of the equation:
A´(x) = 4 - 2*x A´(x) = 0 4 - 2*x = 0
2*x = 4
x = 2 cm
And y = 4 - 2 = 2 cm
The rectangle of maximum area result to be a square of side 2 cm
A(max) = 2*2 = 4 cm²
To find out if A(x) has a maximum in the point x = 2
We get the second derivative
A´´(x) = -2 A´´(x) < 0 then A(x) has a maximum at x = 2
