Actual value approximation? I am having trouble on a calculus problem from an assignment (the problem was removed from the assignment, hence it isn't homework, but I found it interesting enough to warrant my investigation):

Let $f$ be the function with derivative $f'(x)=\sin(x^2)$ and
  $f(0)=-1$.
a) Find the tangent line to $f$ at $x=0$. (I arrived at $y=-1$, which
  is the answer given)
b) Use your answer from part (a) to approximate the value of $f$ at
  $x=0.1$. (I arrived at $-1$, which again is the answer given.)
c) Is the actual value of $f$ at $x=0.1$ greater than or less than the
  approximation from part (b)? Justify your answer.

The only answers given on the document were to part (a) and (b), I have no clue how to even begin on doing part (c).
Note: My calculus teacher has not even touched on anti-derivatives or integration yet.
 A: The question essentially boils down to the following:
Is the first derivative of the function increasing or decreasing? That is, if you 'zoom in' enough, does the function look concave of convex? If it's convex, you can imagine it looks a little like $x^2$, therefore the function will always be greater than any point along its tangent. If it looks concave (e.g. like $-x^2$) then it will always be smaller than any point on its tangent!
Of course, this function isn't concave or convex, unless you restrict yourself to a point near $x=0$. Hence, the question is: is the function's second derivative (also known as its curvature) greater than zero or less than zero at that point.
Note that in this case, $\sin(x^2) \ge 0$ and increases at every point, therefore the function lies above the line you drew (e.g. it is 'locally convex' whenever $x \to 0^+$). Note that this is not the case when $x \to 0^-$!
A: Locally, near $x=0$, we have $\sin(x^2) \geq 0$, with $\sin(x^2)=0$ only when $x=0$. Therefore $f(x)$ is increasing, except at $x=0$, where there is a point of inflection.
The approximating tangent line $y=-1$ therefore lies below the actual value at $x=0.1$.
A: Sometimes a figure is worth a thousand words:

It is easy to find the tangent line at $x=0$:  You find the derivative is $f^\prime (x)|_{x=0} = \sin 0^2 = 0$, and you're given its value there is $-1$.  Hence the equation of the tangent is $f(x) = -1$.
If you take the simple linear approximation, then of course the value at $x = 0.1$ is just $-1$, as shown by the orange line.
To find if the approximation is higher or lower than the actual value, you calculate the second derivative:
${d \sin x^2 \over d x} = 2 x \cos x^2$ which is positive for $x=\epsilon>0$.  Hence the actual curve moves up for values above $x=0$, and the approximation is hence low.
(By the way, the actual function is a Fresnel sine function.)
