Tangent line on curve

Question

Find the points on the curve where the tangent line has a slope of 1.

Answer 1

x^4/4 -3x^3 +2x^2 -3x +5

take the derivative and set it =1

x^3 -x^2 +4x-3=1

solve for x, 3 solutions, graph it, where it intersects the line y=1 are the real solution(s) to the equation and the x coordinates of the points with slope =1. plug those x values into the 4th degree polynomial to find the y coordinates

max 3 positive real solutions, by descartes' method

max zero negative real solutions, i.e. there are no real negative solutions

there are either 3 real positive solutions or 1 real positive and 2 imaginary solutions

as imaginary solutions come in conjugate pairs

solving 3rd degree equations is often trial and error. Try the simplest integers, such as +1

x=1 is one solution,

1/4 -1/3 + 2 -3+5 = 3 11/12

so (1, 47/12)= (1, 3 11/12) is one point with tangent =1

divide x-1 into the 3rd degree polynomial to get a 2nd degree polynomal

use long or synthetic division

set the quotient =0 and solve for x.

x^2+4 = 0 has only 2 imaginary solutions, -2i and +2i

(1, 47/12) is the only point with a tangent line that has slope =1