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generate a continuous and differentiable function f(x) with the following properties: f(x) is decreasing at x=−5 f(x) has a local minimum at x=−3 f(x) has a local maximum at x=3

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mathmate

Answer:

see details in graph and below

Step-by-step explanation:

There are many ways to generate the function.

We'll generate a function whose first derivative f'(x) satisfies the required conditions, say, a quadratic.

1. f(x) has a local minimum at x = -3, and

2. a local maximum at x = 3

Therefore f'(x) has to cross the x-axis at x = -3 and x=+3.

Furthermore, f'(x) must be increasing at x=-3 and decreasing at x=+3.

f'(x) = -x^2+9

will satisfy the above conditions.

Finally f(x) must be decreasing at x= -5, which implies that f'(-5) must be negative.

Check: f'(-5) = -(-5)^2+9 = -25+9 = -16 < 0  so ok.

f(x) can then be obtained by integrating f'(x) :

f(x) = integral of -x^2+9 = -x^3/3 + 9x = 9x - x^3/3

A graph of f(x) is attached, and is found to satisfy all three conditions.

Ver imagen mathmate
MrRoyal

A function is differentiable at [tex]x = a[/tex], if the function is continuous at [tex]x = a[/tex]. The function that satisfy the given properties is [tex]f(x) = 9x - \frac{x^3}{3} + 3[/tex]

Given that:

The function decreases at [tex]x = -5[/tex] means that: [tex]f(-5) < 0[/tex]

The local minimum at [tex]x = -3[/tex] and local maximum at [tex]x = 3[/tex] means that:

[tex]x = -3[/tex] or [tex]x = 3[/tex]

Equate both equations to 0

[tex]x + 3 = 0[/tex] or [tex]3 - x = 0[/tex]

Multiply both equations to give y'

[tex]y' = (3 - x) \times (x + 3)[/tex]

Open bracket

[tex]y' = 3x + 9 - x^2 - 3x[/tex]

Collect like terms

[tex]y' = 3x - 3x+ 9 - x^2[/tex]

[tex]y' = 9 - x^2[/tex]

Integrate y'

[tex]y = \frac{9x^{0+1}}{0+1} - \frac{x^{2+1}}{2+1} + c[/tex]

[tex]y = \frac{9x^1}{1} - \frac{x^3}{3} + c[/tex]

[tex]y = 9x - \frac{x^3}{3} + c[/tex]

Express as a function

[tex]f(x) = 9x - \frac{x^3}{3} + c[/tex]

[tex]f(-5) < 0[/tex] implies that:

[tex]9\times -5 - \frac{(-5)^3}{3} + c < 0[/tex]

[tex]-45 - \frac{-125}{3} + c < 0[/tex]

[tex]-45 + \frac{125}{3} + c < 0[/tex]

Take LCM

[tex]\frac{-135 + 125}{3} + c < 0[/tex]

[tex]-\frac{10}{3} + c < 0[/tex]

Collect like terms

[tex]c < \frac{10}{3}[/tex]

[tex]c <3.33[/tex]

We can then assume the value of c to be

[tex]c=3[/tex] or any other value less than 3.33

Substitute [tex]c=3[/tex] in [tex]f(x) = 9x - \frac{x^3}{3} + c[/tex]

[tex]f(x) = 9x - \frac{x^3}{3} + 3[/tex]

See attachment for the function of f(x)

Read more about continuous and differentiable function at:

https://brainly.com/question/19590547

Ver imagen MrRoyal

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