Question

Exploration Let \(y_{1}=a^{x}, y_{2}=\mathrm{NDER} y_{1}, y_{3}=y_{2} / y_{1},\) and \(y_{4}=e^{y_{3}}\) (a) Describe the graph of \(y_{4}\) for \(a=2,3,4,5 .\) Generalize your description to an arbitrary \(a>1\) (b) Describe the graph of \(y_{3}\) for \(a=2,3,4,\) 5. Compare a table of values for \(y_{3}\) for \(a=2,3,4,5\) with \(\ln a\) . Generalize your description to an arbitrary \(a>1\) (c) Explain how parts (a) and (b) support the statement \(\frac{d}{d x} a^{x}=a^{x} \quad\) if and only if \(\quad a=e\) (d) Show algebraically that \(y_{1}=y_{2}\) if and only if \(a=e\) .

Short answer

From the graph analysis, it can be concluded that the graph of \(y_{3}\) is a straight line with a slope of \(\ln a\), while the graph of \(y_{4}\) is an exponentially growing function. Further, the relationship \(\frac{d}{d x} a^{x}=a^{x}\) holds true only if \(a=e\). Also, the values for \(y_{1}\) and \(y_{2}\) will be equal if and only if \(a=e\). This can be proven algebraically.

Step-by-step solution

Graph of \(y_{4}\)

Plot the given function \(y_{4}=e^{y_{3}}\) for \(a=2,3,4,5\) and observe the shape of the graph for each value of 'a'. Notice that for any positive 'a', the graph shows an exponential growth pattern, with the rate of growth increasing as 'a' increases.

Graph of \(y_{3}\)

Plot the function \(y_{3}=y_{2}/y_{1}\) for \(a=2,3,4,5\). Now observe that regardless of the 'a' value the graph is a straight line with a slope value equal to \(\ln a\). This suggests that \(y_{3}\) is a linear function of x with a slope that depends on the value of 'a'.

Parts (a) and (b) interpretation

Based on the findings from (a) and (b), conclude that the derivative of \(a^{x}\) equals \(a^{x}\) if and only if 'a' is equal to e. This stems from the fact that the functions \(y_{3}\) and \(y_{4}\) only coincide if 'a' is e, as shown in the graphs.

Prove \(y_{1}=y_{2}\) if and only if \(a=e\)

Given \(y_{1}=a^{x}\) and \(y_{2}= \frac{d}{d x} a^{x}\), observe that \(y_{1}=y_{2}\) holds true if and only if \(a=e\). To prove this algebraically, start by finding the derivative of \(y_{1}\) respect to x. Then set \(y_{1}\) equal to \(y_{2}\) and solve the equation to get 'a'. The value will be 'e'.