Calculus of a Single Variable, Chapter 3, 3.2, Section 3.2, Problem 50

Given f(x)=x^4-2x^3+x^2 on the interval [0,6]:
(1) This is a quartic polynomial with positive leading coefficient so its end behavior is the same as a parabola opening up.
(2) f(0)=0 and f(6)=900. Since the function is everywhere continuous and infinitely differentiable everywhere, the Mean Value theorem guarantees the existence of a c in the interval such that the slope of the tangent line at c is the same as the slope of the secant line through the endpoints of the interval.
The slope of the secant line: m=(900-0)/(6-0)=150
The equation of the secant line is y=150x
(3) The derivative of f is 4x^3-6x^2+2x . We set this equal to 150:
x~~3.8721 so y~~123.678 and the equation of the tangent line is:
y-123.678=150(x-3.8721)
The graph of the function, the secant line, and the tangent line: