Answered

Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Join our platform to connect with experts ready to provide precise answers to your questions in different areas. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

On what interval is the functioning decreasing m(x)=4x^3-5x^2-7x?

Sagot :

olemakpadu
m(x) = 4x^3 - 5x^2 - 7x
Let us first find the zeros of the function.
That is when it is equal to zero.

m(x) = 4x^3 - 5x^2 - 7x  = 0
x(4x^2 - 5x - 7) = 0.  Therefore x = 0 or 4x^2 - 5x - 7 = 0.
Using a quadratic function calculator to solve 4x^2 - 5x - 7
x = 2.09, -0.84
Therefore the zeros are x =-0.84, 0, 2.09 for the function m(x).

The intervals observed are imagining  that the zeros are on the number line:
x<-0.84,    -0.84<x<0,  0<x<2.09,  x>2.09.

For each of this range we would test the function with a number that falls in the range.

The function is decreasing in the interval where it is less than 0.
For x<-0.84, let us test x = -1, m(x) = 4x^3 - 5x^2 - 7x = 4(-1)^3 - 5(-1)^2 - 7(-1) = -4 -5 +7 = -2,    -2 < 0, so it is decreasing here.

For  -0.84<x<0, let us test x = -0.5, m(x) = 4x^3 - 5x^2 - 7x = 4(-0.5)^3 - 5(-0.5)^2 - 7(-0.5) = -0.5 -1.25 +3.5 = 1.75,  1.75 >0. It is not decreasing.


For  0<x<2.09, let us test x = 1, m(x) = 4x^3 - 5x^2 - 7x = 4(1)^3 - 5(1)^2 - 7(1) = 4 -5 -7 = -8,  -8 <0. It is decreasing.

For  x>2.09, let us test x = 3, m(x) = 4x^3 - 5x^2 - 7x = 4(3)^3 - 5(3)^2 - 7(3) = 108 -45 -21 = 42,  42 >0. It is not decreasing.

So the function is decreasing in the intervals:

x < -0.84,  & 0<x<2.09.

Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.