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

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
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
For 0
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