A car starts with a dull tank of gas. After driving around 5he city, 1/7 of the gas has been used. With the rest of the gas in the car, the car can travel to and from Ottawa three times. What fractions of a tank of gas does each complete trip to Ottawa use?

Step-by-step explanation:

Given:

A car starts with a dull tank of gas

1/7 of the gas has been used around the city.

With the rest of the gas in the car, the car can travel to and from Ottawa three times.

Question asked:

What fractions of a tank of gas does each complete trip to Ottawa use?

Solution:

Fuel used around the city =

Remaining fuel after driving around the city = 1 -

                                                                         =    

According to question:

As from the rest of the gas in the car that is , the car can complete 3 trip to Ottawa  which means,

By unitary method:

The car can complete 3 trip by using = tank of gas.

The car can complete 1 trip by using =  

                                                             =

                                                             =  

                                                             = tank of gas

Thus, tank of gas used for each complete trip to Ottawa.

(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).