Military specifications often call for electronic devices to be able to withstand accelerations of 10 g. to make sure that their products meet this specification, manufacturers test them using a shaking table that can vibrate a device at various specified frequencies and amplitudes. if a device is given a vibration of amplitude 9.4 cm, what should be its frequency in order to test for compliance with the 10 g military specification? the acceleration of gravity is 9.81 m/s 2 . answer in units of hz.

The solution for this problem is:
(10 x 9.8) = 98.1 m/sec^2 acceleration. Time, to travel 9.4cm or (.094m.), at acceleration of 98m/sec^2

= sqrt(2d/a), = sqrt (98.1 m/sec^2/0.094m) = 32.3050619 sec per cycle

Frequency = (w/2pi), = 32.3050619/2pi
= 32.3050619/6.28318531
= 5.14 Hz would be the answer

(a) Increasing in the interval (-623, 2) and (623, ∞)

f is decreasing in the interval (∞, -623) and (2, - 623).

(b) Local maximum is f = 2 .

2 local minima  f = -623.

Step-by-step explanation:

f(x) = x^4 - 50x^2 + 2

Finding the derivative:

f'(x) = 4x^3 - 100x

This = 0 at  a turning point:

4x(x^2 - 25) = 0

x = 0,  x + -5, 5.

Find the second derivative:

f"(x) = 12x^2 - 100

When x = 0 f(x) = 2

When x = 0, f"(x) is negative so the point (0, 2) is a maximum .

When x = -5, f"(x) is  positive . Also positive when x = 5.

So the 2 local minimums are (-5, (-5)^4 - 50(-5)^2 + 2) = (-5, -623) and

(5, -623).