So, f[x] = 1/4x^2 - 1/2Ln(x)
<span>thus f'[x] = 1/4*2x - 1/2*(1/x) = x/2 - 1/2x </span>
<span>thus f'[x]^2 = (x^2)/4 - 2*(x/2)*(1/2x) + 1/(4x^2) = (x^2)/4 - 1/2 + 1/(4x^2) </span>
<span>thus f'[x]^2 + 1 = (x^2)/4 + 1/2 + 1/(4x^2) = (x/2 + 1/2x)^2 </span>
<span>thus Sqrt[...] = (x/2 + 1/2x) </span>
One of the zeroes would be : x = 1
f (1) = 1^3 + 6(1) ^2 + 3 (1) - 10
f(1) = 1 + 6 + 3 - 10
f (1) = 10 - 10
f (1) = 0
Hope this helps
Answer:
3/4m
Step-by-step explanation:
Answer:
Step-by-step explanation:
Hey there! I hope this helps:
If you have a graphing calculator, go to y=, then enter, and plug each equation in then graph. The graph should show you which equation belongs to which line. You can search "graphing calculator" and plug the equation in and it'll graph it for you
Answer:
it is >
Step-by-step explanation: