F.LE Do two points always determine an exponential function?

1. If we evaluate \(f(x)\) at \(x = 0\) we find $$ f(0) = ab^{0} = ab^0 = a. $$ So the graph of \(f(x) = ab^{x}\) contains \((0,5)\) when \(a = 5\). In order to contain \(Q\) we need \(f(3) = -3\). If \(f(x)\) goes through \((0,5)\) then it is of the form \(f(x) = 5b^{x}\) for some \(b\). When \(x = 3\), \(f\) takes the value \(5b^{3}\). Since \(b\) is the base of an exponential function, it must be positive. So regardless of what value \(b\) is, \(5b^{3}\) is a positive number and so can never be equal to \(-3\). So none of the graphs of the exponential functions passing through \(P\) also pass through \(Q\). Below are graphs of the functions \(f(x)=a\cdot b^x\) with \(a=5\) and for \(b\) equalling the values \(\frac{1}{2}\), \(\frac{2}{3}\), \(1\), \(\frac{3}{2}\), and \(2\).

   
   

2. The graph of \(f(x) = ab^{x}\) passes through \(R = (2,0)\) when \(f(2) = 0\). This is true when $$ ab^{x} = 0. $$ As we saw in part (a), \(b^{x}\) only takes positive values so the only way \(ab^{x}\) could be zero is if \(a = 0\). Once \(a = 0\) then, regardless of what value \(b\) takes, \(f(x) = 0\), a linear function. This is not considered an exponential function and so there is no exponential function whose graph contains the point \(R\).

3. The graph of \(f(x) =ab^{x}\) contains \(S = (c,d)\) when $$ d = ab^{c}. $$ The graph of \(f(x) =ab^{x}\) contains \(T = (g,h)\) when $$ h = ab^{g}. $$ To find \(b\) we use the two equations: \begin{eqnarray*} d &=& ab^{c} \\ h &=& ab^{g} \end{eqnarray*} Dividing the second equation by the first (which is legitimate as neither of these expressions is zero) gives $$ b^{g-c} = \frac{h}{d}. $$ If we take the natural logarithm of both sides of this equation we find $$ (g-c)\ln{b} = \ln{\left(\frac{h}{d}\right)} $$ or $$ \ln{b} = \frac{\ln{\left(\frac{h}{d}\right)}}{g-c} $$ Note that the domain of the natural logarithm is the positive real numbers and this is why we require the hypothesis that \(h\) and \(d\) must either both be positive or both negative. We also need to know that \(g \neq c\) so that dividing by \(g - c\) is legitimate. Finally, we exponentiate both sides of the last equation, with a base of \(e\), in order to solve for \(b\): $$ b = e^{\left(\frac{\ln{\left(\frac{h}{d}\right)}}{g-c} \right)}. $$

   Now that we have found \(b\) we can find \(a\) by plugging the value of \(b\) into one of the two original equations. Note first that if \(d\) and \(h\) are positive then \(a\) is positive while if \(d\) and \(h\) are both negative then \(a\) is negative. We will treat the case here where all three are positive. If we use \(d = ab^{c}\) we find $$ d = ae^{\left(\frac{\ln{\left(\frac{h}{d}\right)}}{g-c} \right)c}. $$ So we can solve for \(a\): $$ a = \frac{d}{e^{\left(\frac{\ln{\left(\frac{h}{d}\right)}}{g-c} \right)c}} $$ Since we have already assumed that \(d\) and \(h\) are positive and \(c \neq g\), these equations for \(a\) and \(b\) make sense and we have found the exponential function of form \(f(x) = ae^{bx}\) whose graph passes through the points \(R\) and \(S\), namely $$ f(x) = ab^x = \frac{d}{e^{\left(\frac{\ln{\left(\frac{h}{d}\right)}}{g-c} \right)c}} e^{\left(\frac{\ln{\left(\frac{h}{d}\right)}}{g-c} \right)x} = de^{\left(\frac{\ln{\left(\frac{h}{d}\right)}}{g-c} \right)(x-c)} $$ If \(d\) and \(h\) are both negative then so is \(a\) and the above equation still makes sense and is correct.

   It is an interesting and worthwhile arithmetic exercise to verify that \(f(x)\) is the correct exponential function, that is that \((c,d)\) and \((g,h)\) lie on its graph. The fact that \((c,d)\) is on the graph of \(y = f(x)\) comes from evaluating the expression \(e^0 = 1\). The fact that \((g,h)\) is on the graph of \(y = f(x)\) comes from the identity $$ e^{\ln{\left(\frac{h}{d}\right)}} = \frac{h}{d}. $$

In preparation for a different approach to part (c), we first resolve part (a) using base \(e\). Since \(f(x)\) is an exponential function, it can be written as $$ f(x) = ae^{kx} $$ for some non-zero number \(k\). (Alternatively, we can define \(k\) as the unique value such that \(e^k=b\).) Evaluating this function \(f(x)\) at \(x = 0\) we find $$ f(0) = ae^{k \cdot 0} = ae^0 = a. $$ So the graph of \(f(x) = ae^{kx}\) contains \((0,5)\) when \(a = 5\) and we can therefore write \(f(x) = 5e^{kx}\). In order to contain \(Q\) we need \(f(3) = -3\). When \(x = 3\), \(f\) takes the value \(5e^{3k}\). Regardless of what value \(k\) is, \(5e^{3k}\) is a positive number and so can never be equal to \(-3\). So none of the graphs of the exponential functions passing through \(P\) also pass through \(Q\).

Part (c) of the problem can be solved by thinking along the lines suggested by part (a). Suppose for the moment that \(c = 0\) as in part (a) of the problem. Then the collection of exponential functions whose graph passes through \(S\) are all of the form $$ f(x) = de^{bx} $$ where \(b\) could be any real number. In order for the graph of this exponential function to pass through \((g,h)\) we must have $$ h = de^{bg} $$ and this allows us to determine \(b\): $$ \ln{h} = \ln{d} + \ln{e^{bg}} = \ln{d} + bg. $$ Solving for \(b\) gives $$ b = \frac{\ln{h} - \ln{d}}{g}. $$ Thus we have found $$ f(x) = de^{\frac{\ln{h} - \ln{d}}{g}x} $$ with the assumption that \(c = 0\).

In general, if \(c\) is not zero, we may first solve the problem, as above, for \(R^\prime = (0,d)\) and \(S^\prime = (g-c,h)\). For this choice of \(R^\prime\) and \(S^\prime\) the exponential function whose graph contains these two points is $$ f(x) = de^{\frac{\ln{h} - \ln{d}}{g-c}x} $$ as we just found above. If we substitute \(x-c\) for \(x\), we find \begin{eqnarray*} g(x) &=& f(x-c) \\ &=& de^{\frac{\ln{h} - \ln{d}}{g-c}(x-c)} \end{eqnarray*} We see that \(g(x)\) is the same as the exponential function \(f(x)\) we found in the first solution, since \(\ln{h} -\ln{d} = \ln\left(\frac{h}{d}\right)\).