The mean value theorem. If f(x) is continuous in [a, b] and differentiable in (a, b), then there exists a point ě in (a, b) such that f(b)- f(a) - f'(E) a <

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Chapter2: Second-order Linear Odes
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6.14) my professor says I have to explain the steps in the solved problems in the picture. Not just copy eveything down from the text.

The mean value theorem. If f(x) is continuous in [a, b] and differentiable in (a, b), then there exists
a point č in (a, b) such that
f(b)- f(a) - f'E) a<5 <b
(16)
b-a
Rolle's theorem is the special case of this where f(a) = f(b) = 0.
The result (16) can be written in various alternative forms; for example, if x and x, are in (a, b),
then
f(x) = f(xo) +f'(Š)(x– x) between x, and x
We can also write result (16) with b = a + h, in which case = a + Oh, where 0< 0 < 1.
(17)
The mean value theorem is also called the law of the mean.
Transcribed Image Text:The mean value theorem. If f(x) is continuous in [a, b] and differentiable in (a, b), then there exists a point č in (a, b) such that f(b)- f(a) - f'E) a<5 <b (16) b-a Rolle's theorem is the special case of this where f(a) = f(b) = 0. The result (16) can be written in various alternative forms; for example, if x and x, are in (a, b), then f(x) = f(xo) +f'(Š)(x– x) between x, and x We can also write result (16) with b = a + h, in which case = a + Oh, where 0< 0 < 1. (17) The mean value theorem is also called the law of the mean.
Differentials
6.14.
Let f(x, y) have continuous first partial derivatives in a region R of the xy plane. Prove that
Af = f(x + Ax, y + Ay) - f(x, y) =f,Ax +f, Ay + 8¡, Ar + 8,Ay
where e, and e, approach zero as Ax and Ay approach zero.
Applying the mean value theorem for functions of one variable (see Page 78), we have
Af = {f(x+ Ax, y + Ay) – f (x, y + Ay)} + {f(x, y + Ay) -f(x, y)}
(1)
= Arf,(x + 0, Ar, y + Ay) + Af,(x, y + 0, Ay)
0 < 0, < 1,0< 0, <1
Since, by hypothesis, f, and f, are continuous, it follows that
f,(x + 0, Ax, y + Ay) = f, (x, y) + 81,
f,(x, y + 0,Ay) = f,(x, y) + d,
where 8, → 0, 8, → 0 as Ax → 0 and Ay → 0.
Thus, Af = f,Ax + f,Ay + 8,Ax + 8,Ay as required.
Defining Ar = dx, Ay = dy, we have Af = f,dx + f, dy + d, dx + d,dy.
We call df = f,dx + f, dy the differential of f (or z) or the principal part of Af (or Az).
Transcribed Image Text:Differentials 6.14. Let f(x, y) have continuous first partial derivatives in a region R of the xy plane. Prove that Af = f(x + Ax, y + Ay) - f(x, y) =f,Ax +f, Ay + 8¡, Ar + 8,Ay where e, and e, approach zero as Ax and Ay approach zero. Applying the mean value theorem for functions of one variable (see Page 78), we have Af = {f(x+ Ax, y + Ay) – f (x, y + Ay)} + {f(x, y + Ay) -f(x, y)} (1) = Arf,(x + 0, Ar, y + Ay) + Af,(x, y + 0, Ay) 0 < 0, < 1,0< 0, <1 Since, by hypothesis, f, and f, are continuous, it follows that f,(x + 0, Ax, y + Ay) = f, (x, y) + 81, f,(x, y + 0,Ay) = f,(x, y) + d, where 8, → 0, 8, → 0 as Ax → 0 and Ay → 0. Thus, Af = f,Ax + f,Ay + 8,Ax + 8,Ay as required. Defining Ar = dx, Ay = dy, we have Af = f,dx + f, dy + d, dx + d,dy. We call df = f,dx + f, dy the differential of f (or z) or the principal part of Af (or Az).
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Let us suppose that the graph of the function f(x) is

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Then the equation of the secant line represented by red in the graph that goes through the two points A and B is

We will now define a new function g(x) as the difference between f(x) and the secant line i.e.

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