Hi there!
c.
Looking at the graph, we can eliminate any possibility of t values on the positive t-axis being an absolute minimum. The only sign change in this interval is from positive to negative, which indicates a maximum.
We can find the value of t = -2 using the integral:
[tex]g(-2) = \int\limits^{-2}_1 {f(t)} \, dt[/tex]
Rewrite in proper format:
[tex]g(-2) = -\int\limits^1_{-2}{f(t)} \, dt[/tex]
Solve using the graph:
[tex]g(-2) = -\frac{1}{2}bh = -\frac{1}{2}(3 \cdot 4) = -6[/tex]
There is an absolute minimum value of g at x = -2. Using the above integral, we get a value of g(-2) = -6, which is the lowest value attainable over the interval.
d.
There is an inflection point at x = 1 because the second derivative changes from positive to negative at this value. (Changes signs). There is NOT an inflection point at x = 2 because the second derivative remains negative, although with a different value.
