Are Slope And Rate Of Change The Same Thing

Feb 4

Why did mathematicians create iii ways of saying ane matter? My estimate is to brand our lives harder, and to take pleasance from mind efffing us! Anyhow I guess it is relevant to know all iii terms and how they relate to each other. So, hither you go!
The Average Rate of Change: The charge per unit of change of a linear function is it's slope! So think = slope formula thou= $\dfrac{y2-y1}{x2-x1}\$ , which is equivalent to the deviation caliber: $\dfrac {f(xth)-f(x)}{h}\$
So taking the divergence quotient is the aforementioned as finding the gradient of a part!
Ex: Discover the divergence quotient for f(x)= 3x+5. Break up the difference quotient into iii steps!
Step one: f(x+h) = 3(ten+h)+5, distribute the 3 and you get: 3x+3h+v
Step 2: f(x+h)- f(x)= 3x+3h+5-(3x+five), hither you distribute the negative and yous come up with 3x+3h+five-3x-5. The +/-3x and +/-5 abolish and you're left with 3h
Step 3: $\dfrac {f(x+h)-f(x)}{h}\$ = $\dfrac{3h}{h}\$ From this fraction y'all can conclude that the h in the numerator and denominator cancel leaving you with merely iii.
So $\dfrac{3(x+h)+5-(3x+5)}{h}\$ =3.
Allow'south not forget most non-linear functions, although I know we all desire to. To find the average rate of change of a non-linear function nosotros need to plot ii points of the role on a graph on an interval of interest. Connect these two points by with a line, called a secant line, the slope of this line will give you the average rate of change.
Ex: Find the difference quotient for f(x)= ${x^2}-3x-2$
Step one: f(x+h)= ${(x+h)^2} +3(x+h)-2$ ,the square is significant considering it requires the (x+h) to multiply by itself resulting in: ${x^2} +2xh+{h^2} +3x+3h-2$
Footstep 2: f(x+h)-f(x)= ${x^2} +2xh+{h^2} +3x+3h-2-({x^2} +3x-2)$ , distribute the negative so that the +/- ${x^2}$ , +/-3x, and +/-2 abolish. Leaving you with $2xh+{h^2}+3h$
Step 3: $\dfrac {f(x+h)-f(x)}{h}$ = $\dfrac{{2xh+{h^2}+3h}}{h}$ Factor a h from the numerator giving you, $\dfrac{h(2x+h+3}{h}$ ,the h cancels from the numerator and denominator giving you 2x+h+3
So $\dfrac {{(x+{h^2})+3(x+h)-2}}{h}= 2x+h+3$
Now find the average charge per unit of change of the coordinate [1,five]! detect h:h =4, because 5-1 equals iv, and 10=ane (the coordinate to the left). Plug these numbers for the variables x and h in the equation 2x+h+three. You get 2(ane)+4+three=ix, therefore the average charge per unit of change equals 9!!!
Now find the deviation quotient of
$f{(x)}=\dfrac{1}{x}$
Step 1: f(x+h) = $\dfrac{1}{x+h}$
Footstep ii: f(x+h)-f(ten)= $\dfrac{1}{x+h}-\dfrac{1}{x}$ , because these fractions have unlike denominators nosotros take to multiply both fractions by opposing fraction denominators to be able to subtract.
$\dfrac{1}{x+h}\times\dfrac{x}{x} = \dfrac{x}{x(x+h}$
$\dfrac{1}{x}\times\dfrac{x+h}{x+h} = \dfrac{x+h}{x(x+h)}$
So f(x+h)-f(x) = $\dfrac{x-(x+h)}{x(x+h)}= \dfrac{x-x-h}{{x^2}+xh}= \dfrac{-h}{{x^2}+xh}$
Footstep three: $\dfrac{f(x+h)-f(x)}{h}$ = $\dfrac{-h}{{x^2}+xh}\div h$ . Call back when dividing y'all accept the reciprocal of the fraction later on the segmentation sign and multiply!
And then it should be written as, $\dfrac{-h}{{x^2}+xh}\times\dfrac{1}{h} = \dfrac{-h}{({x^2}+xh)h}= \dfrac{-1}{{x^2}+xh}$
Employ this deviation quotient to find the slope of the secant line between (2,f(two)) and (7,f(vii)). h equals the difference of seven and 2, so h=5. And for x nosotros use the ten-coordinate in the commencement ordered pair, so ten=ii. Plug in the values for ten and h in the difference quotient $d\frac{-1}{{x^2}+xh}=\dfrac{-1}{{5^2}+(2)(5)}=\dfrac{-1}{14}$
Okay and so nosotros learned three things; (ane) that rational functions suck, peculiarly when trying to find the difference caliber, (2) that the Boilerplate rate of change= Slope of a secant line= Difference Quotient, and (3) if you make a mistake when trying to observe f(x+h) you lot're screwed and everything will come out wrong! And then exist extra careful and double cheque your work!