Unit 5 practice test: Analytical Applications of Differentiation

$f'(x) = \cos x + \frac{1}{2}$ and $f''(x) = -\sin x$. On $(\pi, 2\pi)$, $\sin x < 0$, so $f''(x) > 0$ and the graph is concave up.

• Correct: $(\pi, 2\pi)$: $f''(x) = -\sin x > 0$ when $\sin x < 0$.
• $(0, \pi)$: There $\sin x > 0$, so $f'' < 0$: concave down.
• $\left(\frac{2\pi}{3}, \frac{4\pi}{3}\right)$: That is where $f$ is decreasing ($f' = \cos x + \frac{1}{2} < 0$).
• $\left(\frac{4\pi}{3}, 2\pi\right)$: $f$ is concave up on this piece, but also on $\left(\pi, \frac{4\pi}{3}\right)$.

$f''\left(\frac{\pi}{2}\right) = -1 < 0$: the graph is concave down there, as it is on all of $(0, \pi)$.

• Positive: the graph is rising there from left to right: Rising is about $f'$, not $f''$.
• Zero: the graph is steepest there: $f''$ is zero at $\pi$, not $\frac{\pi}{2}$.
• Correct: Negative: the graph bends down there: $f''\left(\frac{\pi}{2}\right) = -\sin\frac{\pi}{2} = -1$, concave down.
• Positive: the graph bends up there: The graph bends down on $(0, \pi)$.

$f''\left(\frac{3\pi}{2}\right)$$= -\sin\frac{3\pi}{2}$$= 1 > 0$, so $f'$ is increasing there: the graph is concave up.

• Decreasing, because $f$ is concave up there: Concave up goes with $f'$ increasing, not decreasing.
• Correct: Increasing, because $f$ is concave up there: Concave up means $f'' > 0$, so $f'$ is increasing.
• Increasing, because $f\left(\frac{3\pi}{2}\right) > 0$: The sign of $f$ says nothing about $f'$.
• Decreasing, because $f$ has a low point near there: A low point of $f$ is about $f' = 0$; whether $f'$ rises depends on $f''$.

The Extreme Value Theorem needs $h$ continuous on $[0, 4]$. $h$ jumps at $x = 2$ ($\lim_{x \to 2^-} h(x) = 2$ but $h(2) = -1$), so the theorem guarantees nothing.

• Yes, because $[0, 4]$ is a closed interval: A closed interval is half the hypothesis; $h$ must also be continuous on it.
• Yes, because $h$ is defined at each point of $[0, 4]$: Defined is not enough: $h$ jumps from values near 2 down to $-1$.
• No, because $h$ is not differentiable at $x = 0$: Differentiability is not a hypothesis of the Extreme Value Theorem, and $x = 0$ is an endpoint.
• Correct: No, because $h$ is not continuous at $x = 2$: The theorem needs continuity on the closed interval; the jump at 2 breaks it.

The lowest value is $h(2) = -1$. For a maximum, $h(x) = x$ gets as close to 2 as you like as $x \to 2^-$ but never equals 2, and no value is larger, so there is no absolute maximum.

• $h$ has neither, because the Extreme Value Theorem does not apply: When a hypothesis fails, the theorem promises nothing; extrema may still exist. Here the minimum does.
• $h$ has an absolute maximum of $2$, at $x = 2$: $h(2) = -1$, not 2: the open circle at $(2, 2)$ is not on the graph.
• Correct: $h$ has an absolute minimum, $-1$, but no absolute maximum: $h(2) = -1$ is the lowest value. Values near $x = 2$ from the left get close to 2 but never reach it, and every value of $h$ is less than 2.
• $h$ has an absolute maximum of $1$, at the endpoint $x = 4$: $h(1.5) = 1.5$ is larger than $h(4) = 1$, so 1 is not the largest value.

On $[2, 4]$, $h(x) = x - 3$ is continuous (including $h(2) = -1$). The Extreme Value Theorem guarantees both extremes: minimum $h(2) = -1$, maximum $h(4) = 1$.

• Nothing, because $h$ jumps at $x = 2$: The jump is approached from the left of 2, outside $[2, 4]$. On $[2, 4]$ itself $h$ is continuous.
• Correct: $h$ is continuous there, so it has both: maximum $1$, minimum $-1$: On $[2, 4]$, $h(x) = x - 3$ with $h(2) = -1$: no jump inside or at the ends of this interval.
• $h$ has an absolute maximum there but perhaps no absolute minimum: The theorem gives both: an absolute maximum and an absolute minimum.
• $h$ has a critical point in $(2, 4)$ where its extremes occur: $h'(x) = 1$ on $(2, 4)$; there are no critical points. The extremes are at the endpoints.