What is returned by the code: range (20, 100, 20)

Answer:

the author has not considered other points of view.

The continuous and differentiable function where f(x) is decreasing at x = −5 f(x) has a local minimum at x = −2 f(x) has a local maximum at x = 2 is given as: y = 9x - (1/3)x³ + 3.

The continuous function differs from the differentiable function in that the curve obtained is a single unbroken curve in the continuous function.

In contrast, if a function has a derivative, it is said to be differentiable.

It is important to note that a function is differentiable when x is set to a if the function is continuous when x = a.

Given the parameters, we state that

f'(5) < 0; and

x = -5

The local minimum is given as:
x = -3;

the local maximum is given as

x = 3

Thus, x = -3 ; alternatively,

x = 3.  With this scenario, we can equate both to zero.

Hence,

x + 3 = 0;

3-x = 0.

To get y' we must multiply both equations to get:

y' = (3-x)(x + 3)

y'   = 3x + 9 - x² - 3x

Collect like terms to derive:

y' = 3x - 3x + 9 - x²; thus

y' = 9-x²

When y' is integrated, the result is

y = 9x - (x³/3) + c

Recall that

F (-5) < 0

This means that:

9 x -5 - (-5³/3) + c < 0
⇒ -45 + 125/3 + c <0
⇒ -10/3 + c < 0

Collecting like terms we have:
c < 10/3; and

c < 3.33

Substituting C into

f(x) = 9x - x³/3 + c; we have

f(x) = 9x - x³/3 + 3, which is the same as  y = 9x - (1/3)x³ + 3.

Learn more about differentiable functions at:
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