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

1 answer:

4 0

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.

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