Question

What is the local maximum value of the function? g (x)=x^4-5x^2+4

Answer 1

$g(x)=x^4-5x^2+4 \\ g'(x)=4x^3-5 \cdot 2x^1+0=4x^3-10x \\ \\ 4 x^{3} -10x=0 \\ 2x(2x^2-5)=0 \\ 2x^2-5=0~~~~and~~~2x=0 \\ 2x^2=5~~~~~~~~~~~~~~~~~x=0 \\ x^2=2.5 \\ x= \sqrt{2.5} \\ x=1,58 ~~~~~~and ~~~~x= -1.58$
__-____-1,58____+_____0_______-__1,58_____+____>x

Locol max: 0