AP Calculus

Special Focus: The Fundamental
Theorem of Calculus
Questions 4 and 5—Round Two
x
Aha! By the antiderivative part of the Fundamental Theorem, if F( x) = ∫ f ( t) dt,
0
x
then F′ ( x) = f ( x). Likewise, if G( x) = ∫ f ( t) dt,
then G′ ( x) = f ( x). Finally, if
2
x
H( x) = ∫ f ( t) dt,
then H′ ( x) = f ( x). No wonder the three graphs had the same
4
behavior. Their slope functions are the same, and therefore they differ at most by
a constant.
The maximum value of the functions F, G, and H occurs at what x-value(s)?
All three will have a local maximum value at x=2 because that is where the common
slope function, f(x) changes sign from positive to negative. This is the only critical point
on the interval [0,4]; the absolute maximum value must also occur at x=2.
The minimum value of the functions F, G, and H occurs at what x-value(s)?
For each, the only critical value occurs at x=2, where a local maximum occurs.
Therefore, the minimum must occur at an endpoint. From our previous work we know
that functions F and H have a value of 0 at both endpoints and that function G has
a value of –8/3 at both endpoints. Therefore, the absolute minimum value for each
function will occur at x=0 and at x=4.
Functions F, G, and H increase on what interval(s)? Decrease on what interval(s)?
The common slope function is positive on (0,2) and negative on (2,4). Thus each of the
functions F, G, and H increases on [0,2] and decreases on [2,4].
Added Question
Teachers can now ask a question about points of inflections that would not have been
accessible to students earlier.
The graphs of F, G, and H have points of inflection at what x-value(s)?
We know that F′ ( x) = f ( x), G′ ( x) = f ( x), and H′ ( x) = f ( x). Therefore,
F′′ ( x) = G′′ ( x) = H′′ ( x) = f ′( x) .
The slope of f changes sign at x=1 and at x=3, and so the respective second derivatives
change sign at these values. This is why inflection points for all three graphs occur at
x=1 and at x=3.
AP® Calculus: 2006–2007 Workshop Materials 55