​ Can someone help me with this calculus question please?

a)

$f'(x) \text{ is }\left\{\begin{matrix}\text{positive}\\\text{negative}\end{matrix}\right\}\text{ when }f(x)\text{ is }\left\{\begin{matrix}\text{strictly increasing}\\\text{strictly decreasing}\end{matrix}\right\}.$

$\implies f(x) \text{ is strictly increasing on the interval }(-\infty, 1)\cup (3, \infty)\text{ and strictly decreasing on the interval } (1, 3).$

What about when x = 1 or x = 3?

Let ε be an arbitrary constant, infinitesmally small.

$f'(1 -\varepsilon) > 0\text{ and }f'(1 + \varepsilon) < 0$

At x = 1, local maximum of f(x) occurs.

$f'(3 -\varepsilon) < 0\text{ and }f'(3 + \varepsilon) > 0$

At x = 3, local minimum of f(x) occurs.

$f'(x) \text{ is }\left\{\begin{matrix}\text{increasing}\\\text{decreasing}\\\text{constant}\end{matrix}\right\}\text{ when }f(x)\text{ }\left\{\begin{matrix}\text{is convex}\\\text{is concave}\\\text{has a point of inflexion}\end{matrix}\right\}.$

$\implies f(x) \text{ is concave on the interval } (-\infty, 2)\text{ and is convex on the interval }(2, \infty).$

$\text{Point of inflexion of }f(x)\text{ occurs at } x =2.$

b)

$f'(x) \text{ is }\left\{\begin{matrix}\text{increasing}\\\text{decreasing}\\\text{constant}\end{matrix}\right\}\text{ when }f''(x)\text{ is }\left\{\begin{matrix}\text{positive}\\\text{negative}\\\text{0}\end{matrix}\right\}.$

$\implies f''(x) \text{ is negative on the interval } (-\infty, 2)\text{ and is positive on the interval }(2, \infty).$

$f''(2) = 0$