Analytical Applications of Differentiation

$g'(c) = 0$ and $g''(c) < 0$

$g'(c) \neq 0$ and $g''(c) < 0$

$g'(c) = 0$ and $g''(c) > 0$

$g'(c) \neq 0$ and $g''(c) = 0$

The function has an inflection point at that location.

The graph is concave down at that point.

The function has a maximum value at that point.

The graph is concave up at that point.

The function $f'(x)$ is increasing at a decreasing rate on that interval.

The function $f(x)$ is concave up on that interval.

The graph of the function has a horizontal tangent line at some point in the interval.

The function has no points of inflection on that interval.

A continuous function ensures the derivatives also exhibit continuous behavior.

The derivatives must be discontinuous where the original function has jumps.

The derivatives will have vertical asymptotes where the original function is continuous.

The continuity of the original function does not affect the derivatives in any way.

The sign of $f'(x)$ cannot be determined.

$f'(x) < 0$

$f'(x) > 0$

$f'(x) = 0$

At $t = 0$.

At $t = 1$.

There are no values of t where acceleration changes signs.

At $t = -1$.

g(t)'s rate remains constant as t approaches zero from above.

INTO₂CkKky₀(dVZaWtpKShtcwXJvYWNodGVzIHplcm8gZnJvbSBhYm92ZSAuaQMSZWc UCI KYB-hXMgaW5mbGVDdGlvbiBwb2ludCBhdCAkd D05aGFykICh0KSBhcHBybWFjaGVzIHplcm8gZnJvbSBhYm92IC4aaEKSlLSqEBobHNQCwEACxEAcROECTCKRCKRkTQoCRBBNTVcIAwgA, meaning that as t gets closer to zero from positive values, just faster each time.

g(t)'s rate of change increases without bound as t approaches zero from above.

Incorrect implication indicates limitless growth negative direction leading towards infinite decrease instead increase suggested correct answer.

The first derivative of $v(t)$, representing acceleration, equals zero but could change signs before or after this point.

The second derivative of $v(t)$ equals zero indicating an inflection point for velocity.

The value of $v(t)$ itself is necessarily at its maximum or minimum.

The first derivative of $v(t)$ equals zero and must remain constant before and after this point.

Nothing can be concluded regarding extrema or points of inflection without additional information on higher derivatives beyond just $h''$

$h(p) = 0$ indicates that $p$ is definitely where either global or local minima occur since $h''(p) > 0$ suggests upwards concavity around $p$.

$h(p)$ is definitely an inflection point because $h''(p) > 0$ implies a change in concavity there.

$h(p)$ could be an inflection point or nothing particularly noteworthy in terms of maxima/minima because only knowledge about $h''(p)$ isn't sufficient to determine the behavior of $h(p)$.

There will be a hole in the graph of $g''(t)$ when $t=4$.

There will be no graph for $g''(t)$ when $t=4$ since $g'(t)$ doesn't exist there.

There will be a vertical asymptote on the graph of $g''(t)$ when $t=4$.

There will be no change to the graph of $g''(t)$ as it crosses $t=4$ since corners don't affect second derivatives.