This is a differential equations problem. We are to work backwards and determine the function f(x) when given f "(x) and initial values.
<span>f ''(x) = 12x^2 + 6x − 4, when integrated with respect to x, yields:
x^3 x^2
f '(x) = 12------ + 6----- - 4x + C, or 4x^3 + 3x^2 - 4x + C, and
3 2
x^4 x^3 x^2
f(x) = 4------- + 3------- - 4------ + Cx + D, or f(x)=x^4 + x^3 - 2x^2 + Cx + D
4 3 2
Now, because f(0)=5, 5=0^4 + 0^3 -2(0)^2 + C(0) + D, so that D=5.
Determine D in the same manner: Let x=1 and find the value of C.
Then the solution, f(x), is x^4 + x^3 - 2x^2 + Cx + 5. Replace C with this value and then you'll have the desired function f(x).</span>
Answer:
2,5, 6
Step-by-step explanation:
a coefficient is the number placed before the variable in equations and the other numbers, in this case 7 and 4, are called constants
2t^2+1
Subtract t^2 from 3t^2 you get 2t^2
which is 2t^2+1 answer
Answer:
It says not a conic section
Step-by-step explanation:
Yes, 1/4 is bigger than 1/20
