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# ​ Can someone help me with this calculus question please?

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I need help w this qustion

Apr 13, 2020

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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)

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

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

$$f''(2) = 0$$

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Apr 14, 2020
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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
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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
edited by Melody  Apr 14, 2020
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a constant where? on the entire line?

Guest Apr 14, 2020
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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