The derivative of the function $\displaystyle f(x) = x^4 - 2x$ is $\displaystyle f'(x) = 4x^3 -2$.  There is no local maximum for this plot.  Nowhere in the plot of the derivative is there a point where the derivative crosses the x-axis going from positive to negative.

We may want to answer true because we associate concavity with increasing functions, but functions that are decreasing can also be concave up.  The rule of thumb is, if the derivative of a function is increasing on an interval, then the function is concave up.  Here, we are only given the information that the derivative is concave up, but no insight into whether or not it is increasing or decreasing.  So this statement is false.

Remember that critical points of a derivative give us insight into maximum, minimums, and points of inflection.  If the derivative crosses the x-axis going from positive to negative then this tells us that the function has a local maximum at this point.  If the derivative crosses the x-axis going from negative to positive, then this tells us that the function has a local minimum at that point.  Remember the derivative tells us about the rate of change.  If the slope of our function changes from negative to positive, then there must be a small trench that is a minimum.  If the slope of our function changes from positive to negative, then there must be a small hill that is a maximum.

This is true.  For all points of the derivative that are above the x-axis, this is telling us the slope of our function is positive and therefore increasing.  For all points of the derivative below the x-axis, this is telling us the slope of our function is negative and therefore decreasing.

From the description of the derivative we know that a local maximum is at $\displaystyle x = -1$.  Since the derivative is crossing the x-axis going from positive to negative at this point, we assume that the slope of the function $\displaystyle f(x)$ is going from positive to negative creating a local maximum.  At $\displaystyle x = 0$ we are give that the derivative crosses the x-axis going from negative to positive.  This means that our function $\displaystyle f(x)$ is going from decreasing to increasing creating a local minimum.  The graph above meets these criteria.

We know that our function will be concave up when the derivative is increasing and concave down when the derivative is decreasing.  The graph of the derivative is increasing on the interval $\displaystyle (-\infty, 0.9)$ and so $\displaystyle f(x)$ will be concave up on this interval.  The graph of the derivative is decreasing on the interval $\displaystyle (0.9,\infty)$ and so $\displaystyle f(x)$ will be concave down on this interval.

Just because a derivative is increasing does not mean that the function $\displaystyle f(x)$ will be positive.  We know that if the derivative is increasing, then the function must be concave up.  We are given no insight as to where the function starts and no insight as to whether the function is positive or negative.

The derivative of $\displaystyle f(x) = x^2$ is $\displaystyle f'(x) = 2x$.  So the graph of the derivative will have a positive slope of two and pass through the origin.

The local maximums will be the points at which the slope is equal to zero and the slope is in the process of changing from positive to negative.  The local minimums will be the points at which the slope is equal to zero and in the process of changing from negative to positive.  Therefore the maximums are $\displaystyle x = -\frac{3\pi}{2}, \frac{\pi}{2}$ and the minimums are  $\displaystyle -\frac{\pi}{2}, \frac{\3\pi}{2}$.

Points of inflection occur when the graph of the derivative cross the x-axis. If the derivative is going from positive to negative, this tells us the function has a local maximum at this point.  If the derivative is going from negative to positive, then the function has a local minimum at that point.  From this we see that there is a local maximum at $\displaystyle x = -\frac{2}{3}$ and a local minimum at $\displaystyle x = 0$.

The function is increasing when its derivative is positive and decreasing when its derivative is negative.  From this we see that the function is increasing on the intervals $\displaystyle (-\infty, -\frac{2}{3})$and $\displaystyle (0, \infty)$ and decreasing on $\displaystyle (-\frac{2}{3}, 0)$ .

The first derivative graph can also tell us about the concavity of our function.  When the derivative is increasing, the function is concave up.  When the derivative is decreasing, the function is concave down.  From this we see that the function is concave up on the interval $\displaystyle (-\frac{1}{3}, -\infty)$ and concave down on $\displaystyle (-\infty, -\frac{1}{3})$.