Exponential functions and their tangents

Given a>1 and ExpfTanl_1.gif, there exists a unique line y = m x+b which goes through the point ExpfTanl_2.gif and is tangent to the graph of  ExpfTanl_3.gif  at that point.  You learned this in the first course in Calculus.  We will explore this connection between points on the graphs of  exponential functions and their tangent lines in this problem.  

Your tasks are:

Task 1: Given a>1 and ExpfTanl_4.gif calculate m and b as described in the previous cell.  Define the function  ax2mb whose variable is a pair {a,x0} and the output is the corresponding pair {m,b}. What is the range of this function?

Task 2: In Task 1 you determined the range of the function  ax2mb. Given two numbers m and b such that the pair {m,b} is in the range of the function  ax2mb, calculate the corresponding pair {a,x0}. Make this into a function mb2ax. The variable for this function should be a pair {m,b} (of a slope and a y-intercept) and the output should be the corresponding  pair {a,x0}.  

Task 3: Test your functions thoroughly. What is ax2mb[mb2ax[{m,b}]]?  What is mb2ax[ax2mb[{a,x0}]]?  Use this as a basis for testing.

Task 4: Illustrate with plots. Use plots and  Manipulate[] to test your functions.

Response to Task 1

In[185]:=

ExpfTanl_5.gif

Out[185]=

ExpfTanl_6.gif

This is the equation for the tangent line

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ExpfTanl_8.gif

The y-intercept is

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ExpfTanl_10.gif

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ExpfTanl_11.gif

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ExpfTanl_13.gif

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ExpfTanl_15.gif

Since we assume that  a> 1, the exponential functions that we consider are increasing. Therefore their slopes are positive, that is: the range for m is m > 0. The range for b is b<1.  

Response to Task 2

We just need to solve two equations below.

In[192]:=

ExpfTanl_16.gif

ExpfTanl_17.gif

Out[192]=

ExpfTanl_18.gif

I will try to simplify this equation.

In[193]:=

ExpfTanl_19.gif

ExpfTanl_20.gif

Out[193]=

ExpfTanl_21.gif

Mathematica needs a human help here. Fortunately, the second equation above can be simplified by substitution ExpfTanl_22.gif. Then we need to solve b=Y-Y ln(Y). Before trying to solve this equationit is helpful to see the plot of the function Y - Y ln(Y):

In[194]:=

ExpfTanl_23.gif

Out[194]=

ExpfTanl_24.gif

Since we have Y > 1 the above graph tells that the equation b=Y-Yln(Y) always has a unique solution for b. Try to get this solution from Mathematica

In[195]:=

ExpfTanl_25.gif

ExpfTanl_26.gif

Out[195]=

ExpfTanl_27.gif

Since Mathematica gives a warning that some solutions might have been missed, check whether the above function is really the inverse of  Y-Y ln(Y). Plot the functions together.

In[196]:=

ExpfTanl_28.gif

Out[196]=

ExpfTanl_29.gif

In[197]:=

ExpfTanl_30.gif

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ExpfTanl_31.gif

Truly, the function with the red graph is the inverse of the function with the blue graph restricted to the interval (1,+∞).

Now we have the solution for ExpfTanl_32.gif in terms of  b, that is ExpfTanl_33.gif. Substitute this quantity in the equation for m  and solve for a. We can ask Mathematica to do this.

In[198]:=

ExpfTanl_34.gif

Out[198]=

ExpfTanl_35.gif

Now we have the solution for a. The solution for ExpfTanl_36.gif we will get from  ExpfTanl_37.gif. But we need to substitute a  that we found above. Let Mathematica solve for ExpfTanl_38.gif.

In[199]:=

ExpfTanl_39.gif

ExpfTanl_40.gif

Out[199]=

ExpfTanl_41.gif

Let us summarize: We found the following two formulas for a  and  ExpfTanl_42.gif:

In[200]:=

ExpfTanl_43.gif

Out[200]=

ExpfTanl_44.gif

Unfortunately, the formulas that we found for a and ExpfTanl_45.gif are not defined for b=0. Therefore we explore what is happening at b=0.

In[201]:=

ExpfTanl_46.gif

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ExpfTanl_47.gif

We see that there are no problems at b=0. To find the value of the function at b=0 we calculate the limits.

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ExpfTanl_48.gif

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ExpfTanl_49.gif

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ExpfTanl_51.gif

In[204]:=

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ExpfTanl_53.gif

Now we are ready to define the function requested in Task 2.

In[205]:=

ExpfTanl_54.gif

In[207]:=

ExpfTanl_55.gif

Out[207]=

ExpfTanl_56.gif

This is a side note:  What if Mathematica could not find the inverse?  Then we would proceed as follows:  


If Mathematica could not find the inverse, then we would make the inverse using  Interpolation[]

In[208]:=

ExpfTanl_57.gif

In[209]:=

ExpfTanl_58.gif

Out[209]=

ExpfTanl_59.gif

Now explore the inverse that we created and compare it with the inverse which Mathematica found.

In[210]:=

ExpfTanl_60.gif

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ExpfTanl_61.gif

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ExpfTanl_62.gif

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ExpfTanl_63.gif

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Out[213]=

ExpfTanl_67.gif

Response to Task 3

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ExpfTanl_68.gif

Out[214]=

ExpfTanl_69.gif

What is ax2mb[mb2ax[{m,b}]]? The answer: It equals {m,b}.

What is mb2ax[ax2mb[{a,x0}]]?  The answer: It equals {a,x0}.

In[215]:=

ExpfTanl_70.gif

Out[215]=

ExpfTanl_71.gif b==0
ExpfTanl_72.gif True

Mathematica can  not confirm our answer. If we give Mathematica more information, then it might be able to confirm our answer.

In[216]:=

ExpfTanl_73.gif

Out[216]=

ExpfTanl_74.gif

Yes! It does confirm our claim. Now move on to the second answer.

In[217]:=

ExpfTanl_75.gif

Out[217]=

ExpfTanl_76.gif ExpfTanl_77.gif
ExpfTanl_78.gif True

Mathematica can  not confirm our second claim. If we give it more information it might be able to confirm our claim.

In[218]:=

ExpfTanl_79.gif

Out[218]=

ExpfTanl_80.gif ExpfTanl_81.gif
ExpfTanl_82.gif True

No, even with more information it does not confirm it. Next we will confirm it using tables of values.

In[219]:=

ExpfTanl_83.gif

Out[219]=

ExpfTanl_84.gif

In[220]:=

ExpfTanl_85.gif

Out[220]=

ExpfTanl_86.gif

This justifies our second answer.

We can confirm our answers with plots as well:  

In[221]:=

ExpfTanl_87.gif

Out[221]=

ExpfTanl_88.gif

This is quite convincing.

In[222]:=

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Out[222]=

ExpfTanl_90.gif

To get a more convincing picture we restrict the range.

In[223]:=

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ExpfTanl_92.gif

Response to Task 4

In[224]:=

ExpfTanl_93.gif

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In[228]:=

ExpfTanl_98.gif

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ExpfTanl_102.gif

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