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calculate the difference quotients and approximate the slope of the tangent line

Calculate The Difference Quotients And Approximate The Slope Of The Tangent Line class=

Sagot :

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)

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