Answer
a) The function f(x)=$x^3+2x$ is plotted over the interval $-1/2\leq{x}\leq3$ using a computer algebra system (Mathematica) or a graphing calculator.
b)For $x_{0}$=0 ,the difference quotient given by q(h)=$h^2+2$ ,and can be plotted using Mathematica.
c)$\lim\limits_{h \to 0}q(h)=2$
d)The secant line is given by $y=x(h^2+2)$
i )For h=3, y=11x
ii)For h=2, y=6x
iii)For h=1, y=3x
Using Mathematica ,we graph these secant line together with f(x) over the interval $-1/2\leq{x}\leq3$
Work Step by Step
a)The given function $$f(x)=x^3+2x , x_{0}=0->
(1)$$ is plotted over the interval using Mathematica:$$x_{0}-1/2\leq{x}\leq(x_{0}+3)$$
$$-1/2\leq{x}\leq3$$
b)For $x_{0}=0$ ,the difference quotient is$$q(h)=[f(x_{0}+h)-f(x_{0}]/h$$$$=>q(h)=[f(0+h)-f(0)]/h$$$$=>q(h)=[h^3+2h-0]/h$$$$=>q(h)=h^2+2->(2)$$
c)Now ,$$\lim\limits_{h \to 0}q(h)=\lim\limits_{h \to 0}(h^2+2)->[from (2)]$$$$=>\lim\limits_{h \to 0}q(h)=2$$
d)The secant line is given by $$y=f(x_{0})+q.(x-x_{0})$$
For $x_{0}=0$,$$y=f(0)+q.(x-0)$$$$=>y=0+(h^2+2).(x-0)->[from(2)]$$$$=>y=x.(h^2+2)$$
i)For h=3,$$y=x.(3^2+2)$$$$=>y=11x->(3)$$
ii)For h=2,$$y=x.(2^2+2)$$$$=>y=6x->(4)$$
iii)For h=1,$$y=x.(1^2+2)$$$$=>y=3x->(5)$$
Using Mathematica ,we can graph these equations (3),(4) and (5) together with (1) over the interval $-1/2\leq{x}\leq3$
