Let f(x) = x^4+ax^3+bx^2+cx+d, where a,b,c,and d are real numbers. Suppose the graph of y=f(x) intersects the graph of y=2x+3 at x =1,2,3. What is the value of f(0) + f(4)?
+287 It seems there's not enough information to find $f(0)+f(4)$; I need to know the value of $f(x)$ at one more point.
We must have \({x}^{4}+a{x}^{3}+b{x}^{2}+cx+d=2x+3\) at x =1, x=2, and x=3. Substituting we get
a + b + c +d = 4 (1)
8a + 4b + 2c + d = - 9 (2)
27a + 9b + 3c +d = - 72 (3)
Subtracting (1) from (2) and (2) from (3) gives
7a +3b + c = -13 (4)
19a +5b +c = - 63 (5)
Subtracting (4) from (5) results in
12a + 2b = - 50 (6)
We are asked to find
f(0) + f(4)= d + 256 + 64a + 16b + 4c + d
= 256 + 64a + 16b + 4c + 2d (7)
Solving (2) for 2c + d we get
2c + d = - 9 - 8a - -4b (8)
Substituting (8) in 7 gives
f(0) + f(4) = 256 +64a +16b +2(- 9 - 8a - 4b)
= 256 +64a +16b - 18 -16a - 8b
= 238 +48a + 8b
= 238 +4(12a +2b) (9)
Substituting for 12a +2b the value we found in (6) completes the lengthy computations:
f(0) + f(4) = 238 + 4(-50) = 38.
Unless I am in error, no ther value of the function is needed.
