The function $\displaystyle v(t)$ gives you the car's speed at time $\displaystyle t$. Therefore, the fact that $\displaystyle v(0)=0$ means that the car's speed is $\displaystyle 0$ at time $\displaystyle 0$. This is equivalent to saying that the car is not moving at time $\displaystyle t=0$. We have to take the derivative of $\displaystyle v$ to make claims about the acceleration.

Notice that the function $\displaystyle g(t)$ is simply the derivative of $\displaystyle f(t)$ with respect to time. To see this, simply use the power rule on each of the two terms. 

$\displaystyle f'(t) = (2)(5)t + 2 = g(t)$

Therefore, $\displaystyle g(t)$ is the rate at which the car's speed changes, a quantity called acceleration.

$\displaystyle g$ is decreasing on those intervals at which $\displaystyle g'(x)< 0$.

$\displaystyle g(x) = x^{3} -9x^{2}+24x+17$

$\displaystyle g'(x) = 3\cdot x^{3-1} -2\cdot 9x^{2-1}+24+0$

$\displaystyle g'(x) = 3x^{2} -18x+24$

We need to find the values of $\displaystyle x$ for which $\displaystyle g'(x) = 3x^{2} -18x+24 < 0$. To that end, we first solve the equation:

$\displaystyle 3x^{2} -18x+24 =0$

$\displaystyle 3 \left (x^{2} -6x+8 \right ) =0$

$\displaystyle 3 \left (x-2 \right )\left (x-4 \right ) =0$

$\displaystyle x=2\textrm{ or } x = 4$

These are the boundary points, so the intervals we need to check are:

$\displaystyle (-\infty ,2)$, $\displaystyle (2,4)$,  and $\displaystyle (4,\infty)$

We check each interval by substituting an arbitrary value from each for $\displaystyle x$.

$\displaystyle (-\infty ,2)$

Choose $\displaystyle x = 0$

$\displaystyle 3x^{2} -18x+24 =3\cdot 0^{2} -18\cdot 0+24 =24 > 0$

$\displaystyle g$ increases on this interval.

$\displaystyle (2,4)$

Choose $\displaystyle x = 3$

$\displaystyle 3x^{2} -18x+24 =3\cdot 3^{2} -18\cdot 3+24 =27 -54 +24 = -3 < 0$

$\displaystyle g$ decreases on this interval.

$\displaystyle (4,\infty)$

Choose $\displaystyle x = 5$

$\displaystyle 3x^{2} -18x+24 =3\cdot 5^{2} -18\cdot 5+24 = 75 -90 +24 = 9 > 0$

$\displaystyle g$ increases on this interval.

The answer is that $\displaystyle g$ decreases on $\displaystyle (2,4)$.

$\displaystyle f$ is increasing on those intervals at which $\displaystyle f'(x)>0$.

$\displaystyle f \left ( x\right ) = x^{3} - \frac{21}{2} x^{2} +30x -6$

$\displaystyle f' \left ( x\right ) = 3x^{3-1} - 2 \cdot \frac{21}{2} x^{2-1} +30 -0$

$\displaystyle f' \left ( x\right ) = 3x^{2} - 21 x +30$

We need to find the values of $\displaystyle x$ for which $\displaystyle f' \left ( x\right ) = 3x^{2} - 21 x +30 > 0$. To that end, we first solve the equation:

$\displaystyle 3x^{2} - 21 x +30 = 0$

$\displaystyle 3(x^{2} - 7 x +10 )= 0$

$\displaystyle 3(x-2)(x-5)= 0$

$\displaystyle x = 2 \textrm{ or }x = 5$

These are the boundary points, so the intervals we need to check are:

$\displaystyle (-\infty ,2)$, $\displaystyle (2,5)$,  and $\displaystyle (5,\infty)$

We check each interval by substituting an arbitrary value from each for $\displaystyle x$.

$\displaystyle (-\infty ,2)$

Choose $\displaystyle x = 0$

$\displaystyle 3x^{2} - 21 x +30 = 3 \cdot 0^{2} - 21 \cdot 0 +30 = 30 > 0$

$\displaystyle f$ increases on this interval.

$\displaystyle (2,5)$

Choose $\displaystyle x = 3$

$\displaystyle 3x^{2} - 21 x +30 = 3 \cdot 3^{2} - 21 \cdot 3 +30 = 27 -63 +30 = -6 < 0$

$\displaystyle f$ decreases on this interval.

$\displaystyle (5,\infty)$

Choose $\displaystyle x = 6$

$\displaystyle 3x^{2} - 21 x +30 = 3 \cdot 6^{2} - 21 \cdot 6 +30 = 108 -126 + 30=12 > 0$

$\displaystyle f$ increases on this interval.

The answer is that $\displaystyle f$ increases on $\displaystyle (-\infty ,2) \cup (5,\infty)$

To find out where the graph shifts from increasing to decreasing, we need to look at the first derivative. 

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

We're going to treat $\displaystyle 15$ as $\displaystyle 15x^0$ since anything to the zero power is one.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2*-8x^{2-1})+(0*15x^{0-1})$

Notice that $\displaystyle (0*15x^{0-1})=0$ since anything times zero is zero.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2*-8x^{1})$

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=-16x$

If we were to graph $\displaystyle y=-16x$, would the y-value change from positive to negative? Yes. Plug in zero for y and solve for x.

$\displaystyle y=-16x$

$\displaystyle 0=-16x$

$\displaystyle \frac{0}{-16}=x$

$\displaystyle 0=x$

To find out where it shifts from decreasing to increasing, we need to look at the first derivative. The shift will happen where the first derivative goes from a negative value to a positive value.

To find the first derivative for this problem, we can use the power rule. The power rule states that we lower the exponent of each of the variables by one and multiply by that original exponent.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(2*4x^{2-1})+(1*\frac{2}{3}x^{1-1})$

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(2*4x^{1})+(1*\frac{2}{3}x^{0})$

Remember that anything to the zero power is one.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(8x)+(1*\frac{2}{3}(1))$

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=8x+\frac{2}{3}$

Can this equation be negative? Yes. Does it shift from negative to positive? Yes. Therefore, it will shift from negative to positive at the point that $\displaystyle 0=8x+\frac{2}{3}$.

$\displaystyle 0=8x+\frac{2}{3}$

$\displaystyle -\frac{2}{3}=8x$

$\displaystyle \frac{-\frac{2}{3}}{8}=x$

$\displaystyle -\frac{1}{12}=x$

To find out when the function shifts from decreasing to increasing, we look at the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(2*5x^{2-1})+(1*12x^{1-1})$

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(2*5x^{1})+(1*12x^{0})$

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12x^0$

Anything to the zero power is one.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12(1)$

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x+12$

From here, we want to know if there is a point at which graph changes from negative to positive. Plug in zero for y and solve for x.

$\displaystyle y=10x+12$

$\displaystyle 0=10x+12$

$\displaystyle -12=10x$

$\displaystyle \frac{-12}{10}=x$

$\displaystyle \frac{-6}{5}=x$

This is the point where the graph shifts from decreasing to increasing.

First, let's find out if the graph is increasing or decreasing. For that, we need the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

We're going to treat $\displaystyle 15$ as $\displaystyle 15x^0$ since anything to the zero power is one.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2*-8x^{2-1})+(0*15x^{0-1})$

Notice that $\displaystyle (0*15x^{0-1})=0$ since anything times zero is zero.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2*-8x^{1})$

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=-16x$

Plug in our given point for $\displaystyle x$. If the result is positive, the function is increasing. If the result is negative, the function is decreasing.

$\displaystyle y=-16x$

$\displaystyle y=-16(2)$

$\displaystyle y=-32$

Our result is negative, therefore the function is decreasing.

To find the concavity, look at the second derivative. If the function is positive at our given point, it is concave. If the function is negative, it is convex.

To find the second derivative we repeat the process, but using $\displaystyle -16x$ as our expression.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(-16x)=1*-16x^{1-1}$

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(-16x)=-16x^{0}$

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(-16x)=-16$

As you can see, our second derivative is a constant. It doesn't matter what point we plug in for $\displaystyle x$; our output will always be negative. Therefore our graph will always be convex.

Combine our two pieces of information to see that at the given point, the graph is decreasing and convex.

This question asks us to examine the concavity of the function $\displaystyle f(x)= x \cdot e^{x^2}$. We will need to find the second derivative in order to determine where the function is concave upward and downward. Whenever its second derivative is positive, a function is concave upward.

Let us begin by finding the first derivative of f(x). We will need to use the Product Rule. According to the Product Rule, if $\displaystyle f(x)=g(x)\cdot h(x)$, then $\displaystyle f'(x)=g'(x)\cdot h(x)+h'(x)\cdot g(x)$. In this particular problem, let $\displaystyle g(x)= x$ and $\displaystyle h(x)=e^{x^2}$. Applying the Product rule, we get

$\displaystyle f'(x)=\left (\frac{\mathrm{d} }{\mathrm{d} x}[x] \right )\cdot e^{x^2}+\left (\frac{\mathrm{d} }{\mathrm{d} x}[e^{ x^2}] \right )\cdot x$

$\displaystyle f'(x)=1\cdot e^{x^2}+\left (\frac{\mathrm{d} }{\mathrm{d} x}[e^{ x^2}] \right )\cdot x$

In order to evaluate the derivative of $\displaystyle e^{x^2}$, we will need to invoke the Chain Rule. According to the Chain Rule, the derivative of a function in the form $\displaystyle f(x)=g(h(x))$ is given by $\displaystyle f'(x)=g'(h(x))\cdot h'(x)$. In finding the derivative of $\displaystyle e^{x^2}$, we will let $\displaystyle g(x)=e^x$ and $\displaystyle h(x)=x^2$.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}[e^{x^2}]=e^{x^2}\cdot \frac{\mathrm{d} }{\mathrm{d} x}[x^2]=2x\cdot e^{x^2}$

We can now finish finding the derivative of the original function. 

$\displaystyle f'(x)=1\cdot e^{x^2}+\left (\frac{\mathrm{d} }{\mathrm{d} x}[e^{ x^2}] \right )\cdot x$

$\displaystyle =e^{x^2}+(2xe^{x^2})\cdot x=e^{x^2}(2x^2+1)$

To summarize, the first derivative of the funciton $\displaystyle f(x)= x \cdot e^{x^2}$ is $\displaystyle f'(x)=e^{x^2}(2x^2+1)$.

 We need the second derivative in order to examine the concavity of f(x), so we will differentiate one more time. Once again, we will have to use the Product Rule in conjunction with the Chain Rule.

$\displaystyle f''(x)=\frac{\mathrm{d} }{\mathrm{d} x}[f'(x)]=\left (\frac{\mathrm{d} }{\mathrm{d} x}[e^{x^2}] \right )\cdot(2x^2+1)+\left (\frac{\mathrm{d} }{\mathrm{d} x} [ 2x^2+1] \right ) \cdot e^{x^2 }$

$\displaystyle =2xe^ {x^2} \cdot (2x^2+1) + 4x\cdot e^ {x^2}=e^{x^2}\cdot(2x(2x^2+1)+4x)$

$\displaystyle =e^{x^2}\cdot(4x^3+6x)$

In order to find where f(x) is concave upward, we must find where f''(x) > 0.

$\displaystyle f''(x)=e^{x^2}(4x^3+6x) > 0$

In order to solve this inequality, we can divide both sides by $\displaystyle e^{x^2}$. Notice that $\displaystyle e^{x^2}$ is always positive (because e raised to any power will be positive); this means that when we divide both sides of the inequality by $\displaystyle e^{x^2}$, we won't have to flip the sign. (If we divide an inequality by a negative quantity, the sign flips.)

Dividing both sides of the inequality by $\displaystyle e^{x^2}$ gives us

$\displaystyle 4x^3+6x>0$

When solving inequalities with polynomials, we often need to factor.

$\displaystyle 2x(2x^2+3)>0$

Notice now that the expression $\displaystyle 2x^2 + 3$ will always be positive, because the smallest value it can take on is 3, when x is equal to zero. Thus, we can safely divide both sides of the inequality by $\displaystyle 2x^2 + 3$ without having to change the direction of the sign. This leaves us with the inequality

$\displaystyle 2x >0$, which clearly only holds when $\displaystyle x > 0$.

Thus, the second derivative of f''(x) will be positive (and f(x) will be concave up) only when $\displaystyle x > 0$. To represent this using interval notation (as the answer choices specify) we would write this as $\displaystyle (0,\infty )$.

The answer is $\displaystyle (0,\infty )$.

To find out if the function is increasing or decreasing, we need to look at the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(2*5x^{2-1})+(1*12x^{1-1})$

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(2*5x^{1})+(1*12x^{0})$

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12x^0$

Anything to the zero power is one.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12(1)$

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x+12$

Now we plug in our given value and find out if the result is positive or negative. If it is positive, the function is increasing. If it is negative, the function is decreasing.

$\displaystyle y=10x+12$

$\displaystyle y=10(-5)+12$

$\displaystyle y=-50+12$

$\displaystyle y=-38$

Therefore, the function is decreasing.

To find out if it is concave or convex, look at the second derivative. If the result is positive, it is convex. If it is negative, then it is concave.

To find the second derivative, we repeat the process using $\displaystyle 10x+12$ as our expression.

We're going to treat $\displaystyle 12$ as $\displaystyle 12x^0$.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=(1*10x^{1-1})+(0*12x^{0-1})$

Notice that $\displaystyle (0*12x^{0-1})=0$ since anything times zero is zero.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=(1*10x^{1-1})$

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=(10x^{0})$

As stated before, anything to the zero power is one.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=10$

Since we get a positive constant, it doesn't matter where we look on the graph, as our second derivative will always be positive. That means that this graph is going to be convex at our given point.

Therefore, the function is decreasing and convex at our given point.