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Question
Thank you for taking my question. I am homeschooling my son in Calculus, but I am having trouble with one concept. This makes it difficult to teach, naturally. I was hoping a good example would help. Could you please show me your answer to the following question and how you get your answer/your steps?

Given f(x)=x^(2)+2x-3
Identify all the extreme values. Also, find the intervals where the function is increasing and decreasing.

Thank you for your time and assistance,
John

Answer
Hi, John. This is a straightforward problem, especially since the function f(x) is a simple polynomial. The approach is to use the function and its 1st and 2nd derivatives to find its key features, such as its extrema and the intervals over which it is increasing/decreasing. Basically, you will be "sketching out" its shape in the x-y plane. I'll try to explain the process that I used, based on my experience. I highly recommend using some graph paper with x and y axes to sketch out the curve as you go along.

So we have,

y = f(x) = x^2 + 2x -3.

(note that I will use y & f(x) interchangeably in the following).

Starting out, we can find where f(x) crosses the y axis by setting x = 0 and draw a little dot on our plot on the y axis. This value is f(0) = -3, as you can easily see.

Now, the x^2 term, being the highest power of x in the polynomial, tells me the function f(x) will look like a parabola, which will open either up (looking like a valley) or open down (a hill).

"Extreme values" simply means the highest (peaks) and/or lowest points of f(x), i.e., the greatest/least values of y corresponding to specific values of x. A very important concept to apply here is that these points occur where the slope of f(x) is zero, meaning the line tangent to f(x) becomes horizontal. The slope of f(x) is of course

d[f(x)]/dx = 2x + 2

(also written as f'(x) or y')

where I have used the "power rule" for derivatives, which you should be familiar with from your studies so far. Applying the rule for zero slope, we have

f'(x) = 0 => 2x +2 = 0 => x = -1 = value of x where f(x) has an extreme value (greatest or least; we don't know quite yet). Note that, since we are dealing with a parabola, we know that there is only one extreme value to find (higher order polynomials have more extrema). This is good!

On our plot we can now put a little dot along the x axis indicating that the extreme value of f(x) will be either above or below it. So which is it? This is most easily determined by calculating the 2nd derivative of f(x), which tells us how the slope is changing as we vary x. This is

f''(x) = df'/dx = d(2x + 2)/dx = 2.

Thus, the slope of f(x) is always positive, everywhere along the x axis, and in particular at the value x = -1, where the extreme value is.

A positive value of f'' (2nd derivative) means the slope (1st derivative) is increasing as we increase x. At the value of x where we have an extreme value (namely x = -1), the slope is negative to the left of x < -1 (i.e., the original function f(x) decreasing) and positive for x > -1 (f(x) increasing). This is the answer to the question of the intervals, namely

-∞ < x < -1  f(x) decreasing
-1 < x < ∞    f(x) increasing.

In the above, keep in mind that we are talking about 3 entities here, the original function f(x), its 1st derivative f'(x) and its 2nd derivative, f''(x).

Although we have the answer, it is instructive to sketch out a little more what f(x) looks like. The resulting plot will give you a visual confirmation of the answer. We know that the curve of f(x) looks like a valley (like the letter " U ") with its deepest point at x = -1. The y value at this point is found by simply plugging x = -1 into the original function

f(-1) = (-1)^2 + 2・(-1) -3 = 1 -2 -3 = -4.

We can now draw a bit more of the curve f(x) by having it go through y = -4 at x = -1 with each side increasing up towards the x axis. It would be nice to know where the curve crosses the x axis so we can draw it on our plot. This is done by setting f(x) = 0 and finding the values of x (called roots BTW) that solve this equation. So we ask, what values of x solve

f(x) = x^2 + 2 -3 = 0   ?

Since this is a quadratic, we know there are 2 roots. We can find these using the quadratic formula (you should know what it is; see textbook) or, by factoring the polynomial (also hopefully familiar)

f(x) = (x-1)(x+3) =

where we can immediately see that x = 1 and x = -3 are roots. On the plot, we can draw f(x) going through these points on the x axis.

Hope this makes sense and good luck.

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