a) [tex]g=f'[/tex] is continuous, so [tex]f[/tex] is also continuous. This means if we were to integrate [tex]g[/tex], the same constant of integration would apply across its entire domain. Over [tex]0<x<1[/tex], we have [tex]g(x)=2x[/tex]. This means that
[tex]f_{0<x<1}(x)=\displaystyle\int2x\,\mathrm dx=x^2+C[/tex]
For [tex]f[/tex] to be continuous, we need the limit as [tex]x\to1^-[/tex] to match [tex]f(1)=3[/tex]. This means we must have
[tex]\displaystyle\lim_{x\to1}x^2+C=1+C=3\implies C=2[/tex]
Now, over [tex]x<-2[/tex], we have [tex]g(x)=-3[/tex], so [tex]f_{x<-2}(x)=-3x+2[/tex], which means [tex]f(-5)=17[/tex].
b) Integrating over [1, 3] is easy; it's just the area of a 2x2 square. So,
[tex]\displaystyle\int_1^6g(x)=4+\int_3^62(x-4)^2\,\mathrm dx=4+6=10[/tex]
c) [tex]f[/tex] is increasing when [tex]f'=g>0[/tex], and concave upward when [tex]f''=g'>0[/tex], i.e. when [tex]g[/tex] is also increasing.
We have [tex]g>0[/tex] over the intervals [tex]0<x<4[/tex] and [tex]x>4[/tex]. We can additionally see that [tex]g'>0[/tex] only on [tex]0<x<1[/tex] and [tex]x>4[/tex].
d) Inflection points occur when [tex]f''=g'=0[/tex], and at such a point, to either side the sign of the second derivative [tex]f''=g'[/tex] changes. We see this happening at [tex]x=4[/tex], for which [tex]g'=0[/tex], and to the left of [tex]x=4[/tex] we have [tex]g[/tex] decreasing, then increasing along the other side.
We also have [tex]g'=0[/tex] along the interval [tex]-1<x<0[/tex], but even if we were to allow an entire interval as a "site of inflection", we can see that [tex]g'>0[/tex] to either side, so concavity would not change.
