Remember that
the difference quotient is equal to
[tex]\frac{H(x+h)-H(x)}{h}[/tex]
we have
H(x)=(1/3)x^2-2
so
[tex]\frac{\frac{1}{3}(x+h)^2-2-\lbrack\frac{1}{3}x^2-2\rbrack}{h}[/tex][tex]\frac{\frac{1}{3}(x^2+2xh+h^2)-2-\frac{1}{3}x^2+2}{h}[/tex]
Simplify
[tex]\frac{\frac{1}{3}x^2+\frac{2}{3}xh+\frac{1}{3}h^2-2-\frac{1}{3}x^2+2}{h}[/tex][tex]\frac{+\frac{2}{3}xh+\frac{1}{3}h^2}{h}=\frac{2}{3}x+\frac{1}{3}h[/tex]
For h=0.1
[tex]\frac{2}{3}x+\frac{1}{3}(0.1)=\frac{2}{3}x+\frac{1}{30}[/tex]
For h=0.01
[tex]\frac{2}{3}x+\frac{1}{3}(0.01)=\frac{2}{3}x+\frac{1}{300}[/tex]
For h=0.001
[tex]\frac{2}{3}x+\frac{1}{3}(0.001)=\frac{2}{3}x+\frac{1}{3,000}[/tex]
Find out the slope at the point (6,10)
For x=6
[tex]\frac{2}{3}(6)+\frac{1}{3}h=4+\frac{1}{3}h[/tex]
For h=0.1
slope is
4+1/30=4+0.033333=4.033333
For h=0.01
4+1/300=4+0.003333=4.003333
For h=0.001
4+1/3,000=4+0.000333=4.000333
the slope is 4 at point (6,10)
