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nalin [4]
2 years ago
13

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=−2 f(x) has a local maximum at x=2
Computers and Technology
1 answer:
butalik [34]2 years ago
7 0

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.

<h3>What is a continuous and differentiable function?</h3>

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.

<h3>What is the solution to the problem above?</h3>

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:
brainly.com/question/15047295
#SPJ1

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// here is code in c++ to find the approx value of "e".

#include <bits/stdc++.h>

using namespace std;

// function to find factorial of a number

double fact(int n){

double f =1.0;

// if n=0 then return 1

if(n==0)

return 1;

for(int a=1;a<=n;++a)

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}

// driver function

int main()

{

// variable

int n;

double sum=0;

cout<<"enter n:";

// read the value of n

cin>>n;

// Calculate the sum of the series

  for (int x = 0; x <= n; x++)

  {

     sum += 1.0/fact(x);

  }

  // print the approx value of "e"

    cout<<"Approx Value of e is: "<<sum<<endl;

return 0;

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Explanation:

Read the value of "n" from user. Declare and initialize variable "sum" to store the sum of series.Create a function to Calculate the factorial of a given number. Calculate the sum of all the term of the series 1+1/1!+1/2!.....+1/n!.This will be the approx value of "e".

Output:

enter n:12

Approx Value of e is: 2.71828

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