Consider the function F(x)= x³ +½x² −2x +3. Use calc and algebra to find any and all stationary points on the graph of the function.Precalc need help ASAP
Solution:
the above function has more than one root or more than one solution. The derivative of the function represents the slope of the tangent line to the curve
F(x)= x³ +½x² −2x +3, F'(x) = 3(x3-1)+(1/2)(2)(x2-1)-2(x1-1)+0
F'(x)=3x2+x-2 is the slope of the tangent line
setting the slope =0 and solving the quadratic equations, you find all the
critical points of interest
3x2+x-2=0, solve by factoring (3x-2)(x+1)=0, then 3x-2=0, x= 2/3
and x+1=0, x=-1 substituting these values of x in the original equation
you get the corresponding values of F(x) such that;
F(x)= x³ +½x² −2x +3, F(x)=(2/3)3+1/2(2/3)2-2(2/3)+3
= 8/27+4/18-4/3+3 = 8/27+2/9-4/3+3/1 = 59/27 and the first point is
(2/3, 59/27)
F(x)= x³ +½x² −2x +3, F(x)=(-1)3+1/2(-1)2-2(-1)+3
= -1+1/2+2+3 = 9/2 and the 2nd point is ( -1, 9/2)
these two points represents points where the slope of the tangent line to the curve is 0 and could be either maximum or minimum points
You can also find the y intercepts where x = 0 F(x)= x³ +½x² −2x +3, setting x = 0, y= F(x)=3 and the point ( 0,3) is a y intercept setting y = 0, then x intercepts can be found upon solving the equation x³ +½x² −2x +3 =0
