Question

How to find the possible turning points of an equation without graphing?

For example, f(x)=x^3-7x^2+12x, the turning points are 0, 3, & 4.

I need it for f(x)=x^4-10x^2+9 How to find the possible turning points of an equation without graphing?

For example, f(x)=x^3-7x^2+12x, the turning points are 0, 3, & 4.

I need it for f(x)=x^4-10x^2+9 @Mathematics

Answer 1

you can factorize f(x)=x^4-10x^2+9 = (x^2 - 1)(x^2+10)
so, the turning points are , x^2=1, x^2=-10,
the first lead you to x1=1, x2=-1, x3=i*sqrt(10), x4=-i*sqrt(10), with i the imaginary number i^2=-1,
in the first case, also you can factorize, f(x)=x^3-7x^2+12x= x(x^2-7x+12) = x(x-3)(x-4), turning point x1=0, x2=3, x3=4,
all done without graphing