For the function f(x) = 1/4e^-x + e^x, prove that the arc length on any interval has the same value as the area under the curve.

Take an arbitrary interval [a, b], where a < b.

Compute the arc length L of y = f(x) over [a, b] :

Now comptue the area A under the curve y = f(x) over [a, b] :

We have

f (x) = 1/4 e ⁻ˣ + e ˣ   →   f ' (x) = -1/4 e ⁻ˣ + e ˣ

Then

√(1 + (f ' (x))²) = √(1 + (-1/4 e ⁻ˣ + e ˣ)²)

… = √(1 + 1/16 e ⁻²ˣ - 1/2 + e ²ˣ)

… = √(1/16 e ⁻²ˣ + 1/2 + e ²ˣ)

… = 1/4 √(e ⁻²ˣ + 8 + 16e ²ˣ)

… = 1/4 √((e ⁻ˣ + 4 e ˣ)²)

… = 1/4 (e ⁻ˣ + 4 e ˣ)

… = 1/4 e ⁻ˣ + e ˣ

… = f (x)

so both A = L for any choice of interval [a, b].

last option

step-by-step explanation:

given that δdef is right and isoceles,

assuming that side df is the hypotenuse means that de an ef are the other 2 sides that intersect at 90° and are of equal length.

because they are of equal length, we can say that they are congruent

because they intersect at 90°, their slopes must be opposite reciprocals.

only the last option satisfies these conditions.

answer: $34.31

step-by-step explanation:

$11.99 x 3 = $35.97

$35.97 - 10% = $32.37

$32.37 + 6% = $34.31