Solving for b^2+c^2 with Function and Mapping

In summary, the conversation discusses finding the value of b^2 + c^2 for a one-one function, where the approach of monotonicity is used. The conversation also includes a hint to use df/dx and complete the square, and a trigonometric identity is suggested to simplify the expression on the right-hand side. Eventually, the use of polar coordinates helps to solve the problem.
  • #1
abhip
9
0
The function is one-one find the value of b^2+c^2[f(x)=x^3+3x^2+4x+bsinx+ccosxI tried the approach of monotonocity as a one one function will be strictly inc or decreasing in it's domain but I'm not being able to figure out b^2+c^2
 
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  • #2
welcome to pf!

hi abhip! welcome to pf! :smile:

(try using the X2 icon just above the Reply box :wink:)

hint: what is df/dx ? :wink:
 
  • #3
i had tried tat approach of finding df(x)/dx>=0 but can't relate to what has to be found in the question.
 
  • #4
abhip said:
i had tried tat approach of finding df(x)/dx>=0 …

show us what you got :smile:
 
  • #5
3x^2+6x+4>=csinx-bcosx x belongs to real numbers not how to get a relation with b^2 and C^2 sorry I am finding it difficult using the exponent feature
 
  • #6
abhip said:
3x^2+6x+4>=csinx-bcosx

now complete the square :wink:
 
  • #7
but how and what to replace the trigonometric expression on RHS of the inequality with
as for LHS it can be written as 3(x+1)^2 +1
 
  • #8
abhip said:
3(x+1)^2 +1

so the minimum of that is … ? :smile:

(oooh, almost forgot … use one of your trigonometric identities to simplify csinx-bcosx :wink:)
 
  • #9
1... but what about b^2 +c^2 how do i introduce them into the inequality because i have a trigonometric expression on the RHS
 
  • #10
Oh thank you i got it now using polar coordinates...thank you very much...love this forum..it's good that you people are able to solve our doubts it's just impossible getting doubts cleared back at my place as the teachers themselves submit on seeing such doubts...and what i like most is u directly didn't pop the answer it gave me a feeling as if i had been involved in the whole process
 

Related to Solving for b^2+c^2 with Function and Mapping

1. What is the purpose of solving for b^2+c^2 with function and mapping?

The purpose of solving for b^2+c^2 with function and mapping is to find the distance between two points on a coordinate plane. This is commonly known as the Pythagorean Theorem, where b and c represent the lengths of the two sides of a right triangle and the sum of their squares is equal to the hypotenuse squared.

2. How do you use function and mapping to solve for b^2+c^2?

To solve for b^2+c^2 with function and mapping, you will need to use the distance formula: d = √((x2 - x1)^2 + (y2 - y1)^2). This formula represents the distance between two points (x1, y1) and (x2, y2) on a coordinate plane. By substituting the coordinates of the two points into the formula, you can solve for the value of b^2+c^2.

3. Can you provide an example of solving for b^2+c^2 with function and mapping?

Sure, let's say we have two points on a coordinate plane: A(3,4) and B(6,8). Using the distance formula, we can find the distance between these two points: d = √((6 - 3)^2 + (8 - 4)^2) = √(3^2 + 4^2) = √(9 + 16) = √25 = 5. So, the value of b^2+c^2 would be 5^2 = 25.

4. What are the applications of solving for b^2+c^2 with function and mapping?

Solving for b^2+c^2 with function and mapping has many practical applications. It can be used in engineering to calculate distances and dimensions, in physics to determine the magnitude of a vector, in navigation to find the distance between two locations, and in many other fields where measurements and distances are involved.

5. Are there any tips for solving for b^2+c^2 with function and mapping?

One useful tip is to always double-check your calculations and make sure you are using the correct formula. It can also be helpful to draw a diagram to visualize the points and distances involved in the problem. Additionally, practice makes perfect, so the more you solve problems using function and mapping, the easier it will become.

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