Applications of differentiation - the graph of a function and its derivative

Generalization of the mean value theorem
If two functions,  f1(x) and  f2(x) are continuous on a closed interval [a, b] and differentiable between its endpoints have identical derivatives on that interval then they differ in a constant.
That is, if we write  j (x) = f1(x) - f2(x), then its derivative j ' (x) =  f1' (x) - f2' (x) = 0  at every point of the interval, therefore j (x) = constant or
f1(x) - f2 (x) = constant.
This theorem we use in the integral calculus.
Example:   Given is the derivative  f ' (x) =  - x2 - 8x - 12  of a cubic function and one of its roots x = - 1.  Find the cubic and its source function and draw the graph of the cubic.
Solution:    Let factorize given quadratic to find its zeros,
f ' (x) =  - x2 - 8x - 12 = - (x + 6)(x + 2),    x1 = - 6  and  x2 = - 2
and calculate its translation in direction of the x-axis
x0 = - an-1/(n an),    (x0)n = 2 = - (- 8)/(2 · (- 1)) = - 4,   (x0)n = 2 = - 4.
Since, a polynomial function and all its successive derivatives have the same horizontal translation
Then, we calculate coefficient a1 of the cubic
where a1 is the coefficient of the source cubic  fs(x) = a3x3 + a1x   or    fs(x) = (-1/3)x3 + 4x.
Finally we will find the coefficient a0 by plugging given x-intercept (-1, 0) into the cubic
The cubic function and its derivative are shown in the figure below.
Let find translation of the graph of the cubic in direction of the y-axis by plugging  x0 = - 4  into its equation,
Thus, the graph of  f (x) in the above figure is drawn translating the source cubic fs (x) = (-1/3)x3 + 4x in direction of the coordinate axes to the inflection point  I (x0, y0)  or  I (- 4, -3).
To check the source cubic expression mentioned above we should plug the coordinates of translations,
x0 and  y0  into      y + y0 = a3(x + x0)3 + a2(x + x0)2 + a1(x + x0) + a0,
Observe that the problem can be solved almost straight considering we know how find a function whose derivative is given. The function we should find is called an antiderivative.
Therefore, to determine whether a function f (x) is an antiderivative, we simply differentiate it and check if the result is equal to given  f ' (x).
Let denote  f1(x) the cubic we solved and differentiate it
Now, let's do reverse, integrate ( or antidifferentiate)   f1' (x), to return back the cubic
Note that the cubic functions,  f1(x) and  f2(x) have the same derivatives and only differ in the constant term, as was stated by the generalization of the mean value theorem above.
The graph of  f2(x), the cubic missing the constant term, passes through the origin. Therefore, we get its graph by translating the graph of the cubic  f1(x), in the positive direction of the y-axis by 25/3, which is shown in the below figure.
Proof, both functions  f1 (x) and  f2 (x) have the same translation  x0  and the absolute value of the difference ( y0)1 - ( y0)2 of the translations in direction of the y-axis equals 25/3. That is,
Thus, the source function of  f2
Therefore,
Note, that is why we should always add a constant C to an antiderivative function as the substitution for the constant term lost by differentiation.
Let's mention here yet another straight method, that is on some way related to what was just shown above, known as Maclaurin's formula or theorem.
An n-th degree polynomial  Pn (x) = anxn + an - 1xn - 1 + · · ·a2x2 + a1x + a0,  applying Maclaurin's theorem, can be written as
Thus, using Maclaurin's formula we can calculate coefficients of a polynomial by evaluating all its successive derivatives at the origin to the n-th order.
Therefore, we can find the coefficients of the cubic given in the above example, except the last term a0
Since given is,       P3' (x) =  - x2 - 8x - 12     then       P3' (0) =  -12
P3'' (x) =  - 2x - 8,                           P3'' (0) =  - 8
and          P3''' (x) =  - 2,                                  P3''' (0) =  - 2
then by plugging obtained values into above Maclaurin's formula
Cauchy's mean value theorem or generalized mean value theorem
If  f (x) and  g (x) are differentiable in an interval (a, b) and continuous on [a, b], then
or        f ' (c) [g (b) - g (a)] = g' (c) [f (b) -  f (a)],    a < c < b.
To prove the above formula let
j (x) = f (x) [g (b) - g (a)] - g (x) [f (b) -  f (a)]
then, by calculating j (a) and j (b) we get
j (a) = j (b) =  f (a) g (b) - g (a) f (b)
thus, j (x) satisfies Rolle's theorem.
Therefore, there exists at least one point c in the interval (a, b) such that
j' (c) =  f ' (c) [g (b) - g (a)] - g' (c) [f (b) -  f (a)] = 0,
what proves the theorem.
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