Approximate Slope of a Function
We are given the function:
[tex]H(x)=8\ln x+3[/tex]
We will find the approximate value of the slope at (e,11).
It's required to use 3 possible values of the approximation differential h.
Let's use h=0.1 and evaluate the function at x = e + 0.1 = 2.8182818
Compute:
[tex]H(e+0.1)=8\ln 2.8182818+3=11.2890193[/tex]
Compute the difference quotient:
[tex]H^{\prime}=\frac{11.2890193-11}{0.1}=2.890193[/tex]
Now we use h=0.01:
[tex]H(e+0.01)=8\ln 2.728281828+3=11.02937635[/tex]
The difference quotient is:
[tex]H^{\prime}=\frac{11.02937635-11}{0.01}=2.9376353[/tex]
Finally, use h=0.001:
[tex]H(e+0.001)=8\ln 2.719281828+3=11.00294249[/tex][tex]H^{\prime}=\frac{11.00294249-11}{0.001}=2.9424943[/tex]
The last result is the most accurate, thus the slope of the tangent line is 2.94
