Question

7,8,11,12,19,21,24

Find the average rate of change of each function on the interval specified. 5. ༼(x) = ན on [,5] 6. q(x)=r on [4, 2] 7, g(x)= 3r -1 on [3,3] 8 h(x) =5-2r on [2. 4] 10. p0=! - +| on[3] 9. f(0) = 6 +m 1:1.3] Find the average rate of change of each function on the interval specified. Your answers will be expressions involving a parameter (b or h). IL ༼(x) =4x -༡ on [1, bj 12. g(x) • 2r =9 on [4, b] 13. A(x)= 3x+ on [2, 2+h] 14, k(x) = 4x-2 on [3. 3+h] | 15 (༧)a on [9, 9+A] 16 Kx) = on [L. I+h] +4 17. J(x) = 3r on [1, 1+h] 18, t)= 4» on 12, 2+h] 19, ༼(x) = 2r +1 on [r, r+h] 20. gtx)= 3r 2 on k, ར+h] Section 1:3 Rates of Change and Behavior of Graphs For each function graphed, estimate the intervals on which the function is increasing and decreasing 1. པ་ དང་རྒྱལ་ ག་ག་ ་བ་ 2I 23.

Answer 1

the average rate of change of a function over the interval [a,b] is given by

$A=\frac{f(b)-f(a)}{b-a}$

7. given function and interval as

$g(x)=3x^3-1$

we get ,

$f(-3)=3(-3)^3-1=-82$

the average rate of change is

$A=\frac{80-(-82)}{3-(-3)}=\frac{162}{6}=\boldsymbol{27}$

8. given function ,

$h(x)=5-2x^2$

$h(4)=5-2(4)^2=-27$

$h(-2)=5-2(-2)^2=-3$

we get rate of change as

$A=\frac{-27-(-3)}{4-(-2)}=\frac{-24}{6}=\boldsymbol{-4}$

11. given that ,

$f(x)=4x^2-7$

$f(b)=4b^2-7$

$f(1)=4(1)^2-7=-3$

we get average rate of change as

$A=\frac{4b^2-7-(-3)}{b-1}=\frac{4b^2-4}{b-1}=\frac{4(b+1)(b-1)}{b-1}=\boldsymbol{4b+4}$

12. given

$g(x)=2x^2-9$

$g(b)=2b^2-9$

$g(4)=2(4)^2-9=32-9=23$

the average rate of change is

$A=\frac{2b^2-9-23}{b-4}=\frac{2b^2-32}{b-4}=\frac{2(b+4)(b-4)}{b-4}=\boldsymbol{2b+8}$

19. given

$f(x)=2x^2+1$

$f(x+h)=2(x+h)^2+1=2x^2+4xh+2h^2+1$

we get rate of change as

$A=\frac{2x^2+4xh+2h^2+1-2x^2-1}{x+h-x}=\frac{4xh+2h^2}{h}=\boldsymbol{4x+2h}$

the function is said to be increasing if y value increases for an increase in x value and it is said to be decreasing if function value decreases for an increase in x value

21 . in the graph the function value decreases for x increases from negative infinity to -1/2 and from x = 2 to positive infinity ,that is

$decreasing: (-\infty,-1/2)\cup (2,\infty)$

24. looking at graph we see that ,

$decreasing: (-\infty,1)$

$increasing: (1,\infty)$