Given the function y=0.1x^3-0.2x^2+0.3x+1
Find the value of x when y=60.4 without any graphs.

Question

Given the function y=0.1x^3-0.2x^2+0.3x+1
Find the value of x when y=60.4 without any graphs.

campbell_st · Answer

well you have 

$$60.4 = 0.1x^3 -0.2x^2 + 0.3x + 1$$

which becomes 

$$0 = 0.1x^2 - 0.2x^2 + 0.3x - 59.4$$

multiply every term by 10

$$0 = x^3 - 2x^2 + 3x - 594$$

I'd use a bit of guess and check 
x = 10   1000 - 200 + 30 - 594 = 236
x = 8 gives 8^3 -2*8^2 +3*8 - 594 = -186

so the zero is between 10 and 8

try x = 9    9^3 - 2x^2 +3*8 - 594 = 0


so the solution is x = 9

anonymous · Answer

haha, yeah. I was trying to avoid that. I found that there is a cubic formula but thanks for the help anyway. I will give you a medal. :)