This problem is better understood with a given figure. Assuming that the flight is in a perfect northwest direction such that the angle is 45°, therefore I believe I have the correct figure to simulate the situation (see attached).

Now we are asked to find for the value of the hypotenuse (flight speed) given the angle and the side opposite to the angle. In this case, we use the sin function:

sin θ = opposite side / hypotenuse

sin 45 = 68 miles per hr / flight

flight = 68 miles per hr / sin 45

flight = 96.17 miles per hr

6 0

3 years ago

F(x)=<img src="https://tex.z-dn.net/?f=%5Csqrt%7Bx%5E2-4%7D" id="TexFormula1" title="\sqrt{x^2-4}" alt="\sqrt{x^2-4}" align="abs

hammer [34]

Solve for x when √(x ² - 4) = 1 :

√(x ² - 4) = 1

x ² - 4 = 1

x ² = 5

x = ±√5

We're looking at x ≤ 0, so we take the negative square root, x = -√5.

This means f (-√5) = 1, or in terms of the inverse of f, we have f ⁻¹(1) = -√5.

Now apply the inverse function theorem:

If f(a) = b, then (f ⁻¹)'(b) = 1 / f '(a).

We have

f(x) = √(x ² - 4) → f '(x) = x / √(x ² - 4)

So if a = -√5 and b = 1, we get

(f ⁻¹)'(1) = 1 / f ' (-√5)

(f ⁻¹)'(1) = √((-√5)² - 4) / (-√5) = -1/√5

The sign must be negative; see the attached plot, and take note of the negatively-sloped tangent line to the inverse of f at x = 1.

3 0

3 years ago