How to differentiate a function and what is an increasing function?

So I have trouble calculating the following question regarding the use of differentiation.
A curve has equation y = 3x^3 - 6x^2 + 4x + 2
Show that the gradient is never negative.

2 Answers
Andrea S.
Jul 31, 2017

To calculate the derivative of this function you need to apply three rules:

1- The derivative is linear, so:

dy/dx = d/dx (3x^3-6x^2+4x+2) = 3 d/dx (x^3) -6 d/dx (x^2) + 4(d/dx x) + d/dx (2)

2- The power rule stating that:

d/dx (x^n) = nx^(n-1)

3- The derivative of a constant is zero.

Then:

dy/dx = 9x^2-12x+4

so the derivative of y(x) is a second order polynomial that we can factorize:

dy/dx = (3x-2)^2

Thus we can see that:

dy/dx >= 0

Jim G.
Jul 31, 2017

"see explanation"

Explanation:

"to determine if a function f(x) is increasing/decreasing"

• " if "f'(x)>0" then" f(x)" is increasing"

• "if "f'(x)<0" then f(x) is decreasing"

"if y increases as x increases then f(x) is increasing"

"if y decreases as x increases then f(x) is decreasing"

y=3x^3-6x^2+4x+2

rArrdy/dx=9x^2-12x+4=(3x-2)^2

AAx inRR(3x-2)^2>0

•color(white)(x)dy/dx=m_(color(red)"tangent")

rArr" gradient is never negative"

f(x)" is increasing"
graph{3x^3-6x^2+4x+2 [-8.89, 8.89, -4.444, 4.445]}

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