Question

Identify intervals on which the function is increasing, decreasing, or constant. g(x) = 1 - (x - 7)^2

Answer 1

The derivative here is easy enough to find, first expand the equation into:
g(x) = 3 - (x^2 - 10x + 25) = -x^2 + 10x - 22

Take the derivative: g'(x) = -2x +10

Now, we have to find when the equation equals zero. So, we set g'(x) = 0. 0 = -2x + 10, 2x = 10, x = 5. When x is 5, this derivative will be zero, and that is the one and only time when the function is constant. When x is less than 5, you can plug in and see that the derivative will be positive, so the function is increasing when x is less than 5. When x is greater than 5, you can plug in and see that the derivative will be negative, so the function decreasing at x is greater than 5.  Hope  this helped :) have a good day!!!!!!!!!!!!!!!!!!!!!!!!

Answer 2

Answer:

At interval (-∞,0) g(x) is increasing and at interval (0,∞) g(x) is decreasing.

Step-by-step explanation:

We have g(x) = 1 - (x - 7)² = 1 - ( x² - 14x + 49) = 1 - x² + 14x - 49 = - x² + 14x - 48

Let us find out the derivative

        g'(x) = -2x +14

         g'(x) = 0 = -2x +14

           x = 7

  g(7) = - 7² + 14 x 7 - 48 = 1

  So the maximum value of the function is at x = 7, the maximum value is 1

Let us check the value of g(x) at x = 8 and x = 6

   g(6) = - 6² + 14 x 6 - 48 = 0

    g(8) = - 8² + 14 x 8 - 48 = 0

So at interval (-∞,0} g(x) is increasing and at interval [0,∞) g(x) is decreasing.