6.3, #8. The graph of F (x) has a horizontal tangent at x_{1} and x_{3}. F increases from x_{1} to x_{3} and increases also beyond x_{3} because its derivative f(x) is positive there. F is steepest at x_{2} where f has a local maximum.

6.3, #16. The graph of f′′ shows f′′ is greatest at x_{1} and least at x_{5}. Moreover, f′′ ≥ 0 on [x_{1}, x_{3}] and f′′ ≤ 0 on [x_{3}, x_{5}]. So f′ is increasing on [x_{1}, x_{3}], decreasing on [x_{3}, x_{5}], and f′ has greatest value f′(x_{3}) = 0. Moreover, f′ has least value f′(x_{1}) or f′(x_{5}). The picture given with the problem has the following feature: the area A of the region bounded by f′′, the x-axis, and the lines x = x_{1}, x = x_{3} is less than the area B of the region bounded by f′′, the x-axis, and the lines x = x_{3}, x = x_{5}. Since A = f′(x_{3})−f′(x_{1}) and B = f′(x_{3})−f′(x_{5}), it follows that f′(x_{5}) < f′(x_{1}). Thus f′(x_{5}) is the least value of f′. We noted earlier that f′(x_{3}) = 0 is the greatest value of f′. So f′(x) ≤ 0 on [x_{1}, x_{5}] and f is a decreasing function. Thus f has greatest value f(x_{1}) and least value f(x_{5}).

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6.3, #17. It is convenient to separate the three cases: 0 ≤ t ≤ 2, 2 ≤ t ≤ 3, t ≥ 3. For 0 ≤ t ≤ 2, the distance s of the car from the starting point is 50t; at the end of this interval of time, s = 100 miles. For 2 ≤ t ≤ 3, s = 100 − 50(t − 2); at the end of this interval of time, s = 50 miles. Finally for t ≥ 3, s = 50 + 50(t − 3). Thus

s = \{ \begin{matrix}50t & for 0 ≤ t ≤ 2 \\ 100 − 50(t − 2) & for 2 ≤ t ≤ 3 \\ 50 + 50(t − 3) & for t ≥ 3. \end{matrix}

1 2 3 4 50 100 150 _{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}

6.4, #4–6.

∫ (t^{2} + t) dt = \frac{t^{3}}{3} + \frac{t^{2}}{2} + C, ∫ cos t dt = sin t + C, ∫ \sqrt{z} dz = \frac{2}{3}x^{3/2} + C.

6.4, #7–9.

∫ \frac{1}{z} dz = ln z + C, ∫ \frac{1}{t^{2}} dt = −t^{−1} + C, ∫ \frac{1}{z^{3}} dz = −\frac{1}{2}z^{−2} + C.

6.4, #10–12.

\begin{matrix}∫ e^{z} dz = e^{z} + C, ∫ sin t dt = − cos t + C, \\ ∫ (2t^{2} + 3t^{3} + 4t^{4}) dt = \frac{2}{3}t^{3} + \frac{3}{4}t^{4} + \frac{4}{5}t^{5} + C \end{matrix}