Let f(x)= ax^7+bx^3+cx-4 where a, b, and c are constants. If f(-5)=7, then find f(5).

Question

Please show work

Answer 1

There are several ways to think about this. Here is one of them.

f(x) = x³ is an odd function with rotational symmetry about the point (0,0).

That means f(-x) = -f(x).

For example: f(2) = 8, f(-2) = -8. Or, f(-3) = -27, f(3) = -(-27) = 27.

The y-coordinate of the center of rotation (0) is the average of the y-coordinates of f(x) and f(-x).

[8 + (-8)]/2 = 0

Now let's try g(x) = x³ - 4. This too has rotational symmetry, but about the point (0,-4).

g(3) = 23 and g(-3) = -31 and [23 + (-31)]/2 = -4 (the y-coordinate of the center of rotation).

In the posted problem g(x) = ax⁷ + bx³ + cx is an odd function with rotation symmetry about (0,0).

The addition of the -4 constant moves the rotation of symmetry to (0, -4).

That means:

[f(-x) + f(x)]/2 = -4

[f(-5) + f(5)]/2 = -4

[7 + f(5)]/2 = -4

7 + f(5) = -8

f(5) = -15

This Desmos graph depicts the situation and also has a slidable point to show that:

[f(x) + f(-x)]/2 is always -4.

desmos.com/calculator/abmfv1z3f5