Question

The graph of f(x) is given below: 11 godina osno obs 11-10-9-8 -7 -6 -5 4 -3 -2 - 1 2 3 4 5 6 7 8 9 10 11 12 a-Draw a tangent to the given figure for x=9 and explain in a few words what this line tells us about the curve of the function f(x). b. In the interpretation of f19), it is said that the result oives an approximation of the function asx is increased by 4. Why is the relationship approximately? Explain and draw the deviation. Now 2 secount lines between x=9 and y = 6.5 on the one hand and x=9 and x=7 on the other hand. What is the benefit of these templates when discussing the instantaneous rote a change of the function in 8=9?

Answer 1

The diagram after all the lines drawn is:

S. || 6 و 6 الو او و ک ا 4 1 . 2 1 2- - ز- - A - - - - -و- و- - - و- -| | و لم لي لم له له

(a)T is the tangent line. The tangent line at $x=9$ gives how fast or slow the function is changing at that point. It also gives a linear approximation to the function about that point.

(b) Around a neighborhood of $x=9$, say in $[8,10]$, the function is continuous and in $(8,10)$, it is differentiable. Hence, applying Taylor's theorem, we have:

$f(x)=f(9)+f'(9)(x-9)+\cdots$

The function thus can be approximated as $f(x)=f(9)+f'(9)(x-9)$ around $x=9$ by neglecting the higher terms. Since we are neglecting some terms, it's an approximation. The drawing of the derivation is the tangent line, and the interpretation that it is the linear approximation to the function (the $f'(9)$ term).

(c) S1 and S2 are the starting points of the secants (they are so close that they almost merge). When we are discussing the instantaneous rate of change of the function at $x=9$, we check how the rate of the length of the secants change as we come closer and closer to the point in question. The two values have been chosen at a distance of $0.5$ and hence would give a crude approximation to the instantaneous rate of change. As we take smaller and smaller intervals, the approximation gets better.