# Thread: Finding f(-2) and f'(-2), assuming f(x) is..

1. ## Finding f(-2) and f'(-2), assuming f(x) is..

Suppose f(2) = 3 and f'(2) = 1; Find f(-2) and f'(-2), assuming that f(x) is:

a) Even

b) Odd

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My friend and I reasoned purely through our own logic that a) Even would be:

f(-2) = 3
f'(-2) = 1

Because even functions are typically parabolas and therefore they have mirrored points.

I'm not sure if that's actually right, and I don't know what to do for b) Odd.

2. More formally, even functions are symmetric about the y-axis and have the property that f(-x) = f(x) for all x. So f(-2) = f(2) = 3, which you correctly reasoned.

As for the derivative of f, consider $f(x) = x^2$, which you know is an even function. The function is decreasing for x < 0 and increasing for x > 0. Therefore the derivative is negative for x < 0 and positive for x > 0, and thus the derivative can't be even. In fact, the derivative is given by $f'(x) = 2x$ which is in fact an odd function, that is, f'(-x) = -f'(x). So for your problem, f'(-2) = -f'(2) = -1 if f is even.

If f is odd, as in part b, then f' will be even. So you'll have f(-2) = -f(2) = 3 and f'(-2) = f'(2) = 1.

It's relatively straightforward to see why the derivative of an even function is always odd, and vice versa. Just draw an even function and draw the tangent lines at say x = 2 and x = -2. You should notice that the slopes of said tangent lines have the same magnitude, but one tangent line is pointing up and the other pointing down. For an odd function, say $f(x) = x^3$, if you repeat the same experiment you'll notice that the tangent lines are parallel.