08Iapproxgradient

= Rates of Change of Polynomials =


 * Recall **
 * the ** average rate of change ** is the gradient of a line joining two points on the curve
 * the ** instantaneous rate of change ** is the gradient of a tangent at that point.


 * In this exercise, we will use the average rate of change of two points very close together
 * to get an approximate value for the instantaneous rate of change.


 * As you can see from the graph at right
 * Comparing the gradients of AB to AC
 * as the distance between the two points gets smaller,
 * the gradient of the line approaches the gradient of the tangent at A
 * Example 1 **

... ... Sketch y = 2x 2.

... ... Find the gradient of a line joining ... ... ... ** (a) ** .. 2 and 3 ... ... ... ** (b) ** .. 2 and 2.1 ... ... ... ** (c) ** .. 2 and 2.01 ... ... ... ** (d) ** .. 2 and 2.001

... ... Use these answers to predict the gradient of a tangent to the curve at x = 2


 * Solution:**

... ... The gradient of a line joining points where x = 2 and x = 3 on the curve is:

math . \qquad \qquad x = 2 \quad \Rightarrow \quad y = 8 \\.\\ . \qquad \qquad x = 3 \quad \Rightarrow \quad y = 18 \\.\\ . \qquad \qquad m = \dfrac{18-8}{3-2} \\.\\ . \qquad \qquad m = \dfrac{10}{1} \\.\\ . \qquad \qquad m = 10 math

... ... In a similar way

... ... ... ** (a) ** .. 2 and 3 ... ... ... m = 10 ... ... ... ** (b) ** .. 2 and 2.1 ... .... m = 8.2 ... ... ... ** (c) ** .. 2 and 2.01 ... ... m = 8.02 ... ... ... ** (d) ** .. 2 and 2.001 ..... m = 8.002

... ... ... We can see that as the distance between the two points gets smaller ... ... ... The gradient is getting closer and closer to m = 8

... ... ... Therefore, we predict that at x = 2, we predict the gradient of the tangent will be m = 8


 * Hybrid Functions **


 * Recall that a hybrid function has different rules for each section of the domain.

If the function is ** discontinuous ** (has a jump) between the two parts of the function
 * the gradient of the hybrid function is __**undefined**__ at that point
 * because we can't draw a tangent to the curve at that point


 * for example,

math . \qquad g \big( x \big) = \left\{ \begin{matrix} x + 2 && x \leqslant 0 \\ x^2 && x>0 \end{matrix} \right. math


 * at x = 0,
 * the left rule (y = x + 2) gives (0, 2)
 * the right rule (y = x 2 ) gives (0, 0)
 * graph is not continuous so gradient of g(x) is not defined at x = 0



If the hybrid function is ** continuous ** but has a ** sharp change of direction ** at the join between the two parts
 * the gradient of the hybrid function is __**undefined**__ at that point
 * because we can't draw a single tangent to the curve at that point (we can draw many lines that touch the sharp point)
 * the gradient at that point according to one of the rules will be __**different**__ to the gradient at that point according to the other rule


 * for example,

math . \qquad h \big( x \big) = \left\{ \begin{matrix} \big( x - 2 \big)^2 && x \leqslant 2 \\ x-1 && x>2 \end{matrix} \right. math


 * at x = 2
 * both rules give (2, 1)
 * the left rule ( y = (x – 3) 2 ) gives m = –2
 * the right rule ( y = x – 1 ) gives m = 1
 * rules give different values so gradient of h(x) is not defined at x = 2

If the hybrid function is ** continuous ** and ** smooth ** at the join then
 * the gradient of the function __**is defined**__ at that point
 * the gradient at that point according to one of the rules will be __**the same**__ as the gradient at that point according to the other rule.


 * for example

math . \qquad f \big( x \big) = \left\{ \begin{matrix} x^2 && x < 1 \\ 2x-1 && x \geqslant 1 \end{matrix} \right. math


 * at x = 1
 * both rules give (1, 1)
 * the left rule ( y = x 2 ) gives m = 2
 * the right rule ( y = 2x – 1 ) gives m = 2
 * rules give the same value so gradient of f(x) is 2 at x = 1

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