#1**+2 **

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)

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

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

\(f''(2) = 0\)

.MaxWong Apr 14, 2020

#2**+1 **

what do you mean by f' is constant when f has a point of inflexion?

Guest Apr 14, 2020

edited by
Guest
Apr 14, 2020

#3**+1 **

If the gradient is not increasing and not decreasing then it must be a constant.

At a point of inflection

\(\frac{d^2y}{dx^2}=0\\~ \\and\\~\\ \frac{dy}{dx}=\int \frac{d^2y}{dx^2}dx=\int 0dx=k \)

**I am not sure that I have presented this properly.**

**Feel free to comment Max.**

Melody
Apr 14, 2020

#5**+1 **

No just at that one point. At the point of inflextion the gradient is neither increasing nor decreasing.

If ithe rate of change is not + or - then it must be 0. Which means that the gradient must be a constant at that point.

It is not something I would usually add. (I can see the confusion)

I would normally have just left the statement out altogether.

It is true though.

Melody
Apr 14, 2020