Evaluating Exponential Functions at a Point
We’ve seen other functions that are exponential expressions before, such as \(f(x)=x^2\) or \(g(x)=x^4\). In those cases, the base of the expression was the variable. In exponential functions, however, the exponent is the variable. Thus, when evaluating the output of such functions for some given input, that’s where the input goes when you substitute!
Replace the \(x\) in the exponent with the given inputs:
\(f(2)=2^2=4\)
\(f(4)=2^4=16\)
Exponential functions can be further generalized by allowing the exponent to be an expression in a variable rather than just a simple variable, like in the above example. This is illustrated in the following example. The process remains the same as any other function-evaluation problem you’ve seen so far: replace variable with the input.
Replace the \(x\) in the exponent with the given inputs:
\(f(2)=2^{2^2+1}=2^{5}=32\)
\(f(4)=2^{4^2+1}=2^{17}=131072\)
\(f(2)=16\)
\(f(3)=64\)
\(f(1/2)=2\)
\(f(2)=4^2=16\)
\(f(3)=4^3=64\)
\(f(1/2)=4^{\frac{1}{2}}=\sqrt{4}=2\)
\(f(2)=\frac{1}{4}\)
\(f(-2)=4\)
\(f(1/2)=\frac{1}{\sqrt{2}}\)
\(f(2)=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)
\(f(-2)=\left(\frac{1}{2}\right)^{-2}=\frac{1}{2^{-2}}=2^2=4\)
\(f(1/2)=\left(\frac{1}{2}\right)^{\frac{1}{2}}=\frac{1}{2^{\frac{1}{2}}}=\frac{1}{\sqrt{2}}\)
\(g(1)=\frac{1}{2}\)
\(g(-1)=32\)
\(g(1)=2^{1^2-3\cdot 1+1}=2^{-1}=\frac{1}{2}\)
\(g(-1)=2^{(-1)^2-3\cdot (-1)+1}=2^{1+3+1}=2^5=32\)
\(f(a)=3\cdot 2^{a}\)
\(f(3a+1)=3\cdot 2^{3a+1}\)
\(f(x^2-1)=3\cdot 2^{x^2-1}\)
With all three of these, the idea is the same: every place you see an \(x\) in the rule for \(f\), replace it with what’s in the parentheses of \(f(\ \ \ \ )\).
In the first, case, we are replacing all \(x\)’s in the rule for \(f\) with \(a\). Hence,
\(f(a)=3\cdot 2^a\)
In the next case, for \(f(3a+1)\), we replace all \(x\)’s in the rule for \(f\) with the expression \((3a+1)\). Hence, \(f(3a+1)=3\cdot 2^{(3a+1)}\). You can safely drop the parentheses if you like since we are not doing anything to the quantity as a whole.
Lastly, for \(f(x^2-1)\) we are replacing every \(x\) in the rule for \(f\) with the expression \((x^2-1)\). Thus, we get \(f(x^2-1)=3\cdot ^{x^2-1}\).
In each of the three problems above, there is not much we can do to simplify, but exponent rules give you a bit you could do to expand / make things uglier.
\(h(1)=1\).
\(h(x+1)=3^{x^2}\).
Do essentially the same thing to find \(h(x^2+1)\)
\(h(x^2+1)=3^{x^4}\)
\(h(1)=3^{(1-1)^2}=3^{0^2}=3^0=1\).
To find \(h(x+1)\), replace every \(x\) you see in the rule for \(h\) with an \((x+1)\) and simplify
\(h(x+1)=3^{((x+1)-1)^2}=3^{x^2}\).
Do essentially the same thing to find \(h(x^2+1)\)
\(h(x^2+1)=3^{((x^2+1)-1)^2}=3^{(x^2)^2}=3^{x^4}\)