Question

f(x)=x^4-50x^2+2

Answer 1

Suppose x^2 = y

Then,

y^2 - 50y + 2
Solve.

Answer 2

i need inflection points

Answer 3

f(x)=x^4-50x^2+2
f'(x)=4x^3 -100x
f''(x) = 12x^2 -100
f''(x)=0
12x^2 -100=0
12x^2 =100
x^2 =25/3
x=sqrt (25/3) or -sqrt (25/3)

「Reference:
for x< -sqrt (25/3), f''(x) >0 concave up
for -sqrt (25/3)<x<sqrt (25/3), f''(x)<0 concave down
for x>sqrt (25/3), f''(x) >0 concave up」

therefore,
pts of inflexion
put x=sqrt (25/3) into f(x), then you can get the y-coord.
put x=-sqrt (25/3) into f(x), then you can get the y-coord.

then, pts of inflxion are found

Answer 4

thank you but somehow i screw it up when i plug sqrt(25/3) into f(x)

Answer 5

sqrt(25/3)^4-50 sqrt(25/3)^2+2
=(25/3)^2 -50(25/3)+2
=-3107/9

Answer 6

thank you thank you!!

Answer 7

you are welcome