We have $6x + 24 = 0$ so $x = -4$. This value breaks the real number line into two intervals, $(-\infty , -4)$ and $(-4, \infty )$. The second derivative maintains the same sign throughout each of these intervals. To determine whether it is positive or negative, we choose a test point in each interval. For the interval, $(-\infty , -4)$, we choose $x = -5$. Plugging this into the second derivative, we get $f''(-5) = 6(-5) + 24 = -6 < 0$. Next, we choose the test point $x = 0$ for the interval $(-4, \infty )$. Plugging this into the second derivative gives $f''(0) = 6(0) + 24 = 24 > 0$. We can use this information to create a sign chart for the second derivative, as shown below.

We can now use the concavity theorem to conclude that the original function, $f(x) = x^3 + 12x^2 - 7x +10$ is concave down on the interval $(-\infty , -4)$. and concave up on the interval $(-4, \infty )$. Finally, noting that the original polynomial is continuous everywhere and changes concavity at $x = -4$, we can say that $f(x) = x^3 + 12x^2 - 7x +10$ has an inflection point at $x = -4$.

We have $12x + 6 = 0$ so $x = -1/2$. This value breaks the real number line into two intervals, $(-\infty , -1/2)$ and $(-1/2, \infty )$. The second derivative maintains the same sign throughout each of these intervals. To determine whether it is positive or negative, we choose a test point in each interval. For the interval, $(-\infty , -1/2)$, we choose $x = -1$. Plugging this into the second derivative, we get $f''(-1) = 12(-1) + 6 = -6 < 0$.

Next, for the interval $(-1/2, \infty )$, we choose the test point $x = 0$. Plugging this into the second derivative gives $f''(0) = 12(0) + 6 = 6 > 0$.

We can now use the concavity theorem to conclude that the original function, $f(x) = 2x^3 + 3x^2 - 36x +2$ is concave down on the interval $(-\infty , -\frac 12)$, and concave up on the interval $(-\frac 12, \infty )$. Finally, noting that the original polynomial is continuous everywhere and changes concavity at $x = -\frac 12$, we can say that $f(x) = 2x^3 + 3x^2 - 36x +2$ has an inflection point at $x = -\frac 12$.

These two values break the real number line into three intervals: $(-\infty , \alpha ), (\alpha , \beta )$ and $(\beta , \infty )$ with test points $x = -2, x = -1$ and $x = 0$, respectively. Plugging these into the second derivative gives $$f''(-2) = (9(-2)^2 + 12(-2) + 2)e^{3(-2)} = 14e^{-6} > 0,$$ $$f''(-1) = (9(-1)^2 + 12(-1) + 2)e^{3(-1)} = -e^{-3} < 0,$$ and $$f''(0) = (9(0)^2 + 12(0) + 2)e^{3(0)} = 2 > 0.$$

We can now use the concavity theorem to conclude that the original function, $f(x) = x^2 e^{3x}$ is concave up on the intervals $(-\infty , \alpha )$ and $(\beta , \infty )$ and it is concave down on the interval $(\alpha , \beta )$. Finally, we can infer from this that the continuous function $f(x) = x^2 e^{3x}$ has inflection points at $$x = \alpha = \dfrac {-2 - \sqrt {2}}{3} \quad \text {and} \quad x = \beta = \dfrac {-2 + \sqrt {2}}{3}.$$

These two values of $x$ break the real number line into three intervals: $(-\infty , -\tfrac {1}{\sqrt 2}), (-\tfrac {1}{\sqrt 2}, \tfrac {1}{\sqrt 2})$ and $(\tfrac {1}{\sqrt 2}, \infty )$. The sign of $f''(x)$ will remain the same on each of these intervals. To determine the sign on each interval, we use the test points $x = -1, x = 0$ and $x = 1$ respectively. Plugging the test points into the second derivative gives $$f''(-1) = (4(-1)^2 - 2) e^{-(-1)^2} = 2e^{-1} > 0$$ $$f''(0) = (4(0)^2 - 2) e^{-(0)^2} = -2 < 0$$ and $$f''(1) = (4(1)^2 - 2) e^{-(1)^2} = 2e^{-1} > 0.$$

We can now use the concavity theorem to conclude that the original function, $f(x) = e^{-x^2}$ is concave up on the intervals $(-\infty , -\frac {1}{\sqrt 2})$ and $(\frac {1}{\sqrt 2}, \infty )$ and concave down on the interval $(-\frac {1}{\sqrt 2},\frac {1}{\sqrt 2})$. Finally, since the original function is continuous everywhere, we can say that $x = \pm \frac {1}{\sqrt 2}$ are inflection points for $f(x) = e^{-x^2}$.

To answer the first question, recall that when $f'(x) > 0$, then $f(x)$ is increasing, and when $f'(x) < 0$, then $f(x)$ is decreasing. From the graph, we see that $f'(x)$ is positive on the intervals $(-\infty , -2)$ and $(2, \infty )$. Hence, the graph of $f(x)$ is increasing on these intervals. We can also see that $f'(x)$ is negative on the interval $(-2, 2)$ and therefore $f(x)$ is decreasing on this interval. We can now use the first derivative test to determine the nature of the local extremes. Since $f'(-2) = 0$ and $f'(x)$ changes sign from positive to negative at $x = -2$, the function, $f(x)$, has a local maximum at $x = -2$. Similarly, Since $f'(2) = 0$ and $f'(x)$ changes sign from negative to positive at $x = 2$, the function, $f(x)$, has a local minimum at $x = 2$.
To determine the concavity of $f(x)$,recall that $f(x)$ is concave up when $f'(x)$ is increasing and $f(x)$ is concave down when $f'(x)$ is decreasing. From the graph, we see that $f'(x)$ is increasing on the interval $(0, \infty )$, and decreasing on the interval $(-\infty , 0)$. Hence, the graph of $f(x)$ is concave up on $(0, \infty )$ and concave down on $(-\infty , 0)$. Finally, $f(x)$ has an inflection point at $x = 0$ due to the change in concavity there.

To answer the first question, recall that when $f'(x) > 0$, then $f(x)$ is increasing, and when $f'(x) < 0$, then $f(x)$ is decreasing. From the graph, we see that $f'(x)$ is positive on the intervals $(0, 2)$ and $(2, \infty )$. Hence, the graph of $f(x)$ is increasing on these intervals. We can also see that $f'(x)$ is negative on the interval $(-\infty , 0)$ and therefore $f(x)$ is decreasing on this interval. We can now use the first derivative test to determine the nature of the local extremes. Since $f'(0) = 0$ and $f'(x)$ changes sign from negative to positive at $x = 0$, the function, $f(x)$, has a local minimum at $x = 0$. The situation at $x = 2$ is different. From the graph, we see that $f'(2) = 0$ but $f'(x)$ does not change sign at $x = 2$- it is positive on either side. Thus, $f(x)$ does not have a local extreme at the critical number, $x = 2$.
To determine the concavity of $f(x)$,recall that $f(x)$ is concave up when $f'(x)$ is increasing and $f(x)$ is concave down when $f'(x)$ is decreasing. From the graph, we see that $f'(x)$ is increasing on the intervals $(-\infty , .7)$ and $(2, \infty )$ and decreasing on the interval $(0.7, 2)$. Hence, the graph of $f(x)$ is concave up on $(-\infty , .7)$ and $(2, \infty )$ and concave down on $(.7, 2)$. Finally, $f(x)$ has an inflection points at $x = .7$ and $x = 2$ due to the changes in concavity at these points.