scalar MVT

Consider scalar functions f:R^n-->R. Am I correct that for functions
like this, a "usual" MVT exists, but not for vector-valued functions?
I think I have a proof that:

f(y) - f(x) = grad(f)(z) dot (y-x) for some z on the line segment
between x and y.

Is this true?

Geometrically, does this just say the following (say f:R^2-->R): If x
and y are two points on the XY plane. Then there exists a z between x
and y such that the directional derivative at z in the direction u = y
- x equals f(y) - f(x)? In other words, the directional derivative
equals the "secant line" at z?

.

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