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.
What is a continuous and differentiable function?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.
What is the solution to the problem above?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.
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