Calculus: Early Transcendentals, Chapter 3, 3.1, Section 3.1, Problem 39

The differential of a polynomial term like Ax^n is nAx^(n-1)
If f(x)=x^4-2x^3+x^2
then f'(x)=(4)(1)x^(4-1)-(3)(2)x^(3-1)+(2)(1)x^(2-1)
Completing the basic math we get f'(x)=4x^3-6x^2+2x
Looking at the graphs of the original function (black) and it's derivative (blue), we see:

For X values less than 0, the slope of the original function (black) is negative (as x values increase towards 0 from the left, the function decreases in value), therefore it's derivative (blue) is negative.
From x=0 to x=0.5, the slope of the original function becomes positive (the value of the function rises as x increases), therefore the derivative (blue) is positive for this duration.
From x=0.5 to x=1, the slope of the original function (black) is again negative as shown by the derivative (blue) again being below the x axis.
For x values above 1, the original function again has a positive slope, resulting in a positive derivative.

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