`f(x) = x^4 - 2x^3 + x^2, [0,6]` Use a graphing utility to (a) graph the function `f` on the given interval, (b) find and graph the secant line through points on the graph of `f` at the endpoints of the given interval, and (c) find and graph any tangent lines to the graph of `f` that are parallel to the secant line.
1 Answer | Add Yours
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:
The graph of the function, the secant line, and the tangent line:
We’ve answered 319,630 questions. We can answer yours, too.Ask a question