Exponential functions and their tangents
Given a>1 and , there exists a unique line y = m x+b which goes through the point and is tangent to the graph of 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 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 yintercept) 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
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This is the equation for the tangent line
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The yintercept is
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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.
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I will try to simplify this equation.
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Mathematica needs a human help here. Fortunately, the second equation above can be simplified by substitution . Then we need to solve b=YY ln(Y). Before trying to solve this equationit is helpful to see the plot of the function Y  Y ln(Y):
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Since we have Y > 1 the above graph tells that the equation b=YYln(Y) always has a unique solution for b. Try to get this solution from Mathematica
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Since Mathematica gives a warning that some solutions might have been missed, check whether the above function is really the inverse of YY ln(Y). Plot the functions together.
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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 in terms of b, that is . Substitute this quantity in the equation for m and solve for a. We can ask Mathematica to do this.
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Now we have the solution for a. The solution for we will get from . But we need to substitute a that we found above. Let Mathematica solve for .
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Let us summarize: We found the following two formulas for a and :
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Unfortunately, the formulas that we found for a and are not defined for b=0. Therefore we explore what is happening at b=0.
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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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Now we are ready to define the function requested in Task 2.
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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[]
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Now explore the inverse that we created and compare it with the inverse which Mathematica found.
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Response to Task 3
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What is ax2mb[mb2ax[{m,b}]]? The answer: It equals {m,b}.
What is mb2ax[ax2mb[{a,x0}]]? The answer: It equals {a,x0}.
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Mathematica can not confirm our answer. If we give Mathematica more information, then it might be able to confirm our answer.
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Yes! It does confirm our claim. Now move on to the second answer.
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Mathematica can not confirm our second claim. If we give it more information it might be able to confirm our claim.
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No, even with more information it does not confirm it. Next we will confirm it using tables of values.
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This justifies our second answer.
We can confirm our answers with plots as well:
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This is quite convincing.
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To get a more convincing picture we restrict the range.
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Response to Task 4
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