Question

given the function below, fill in the table of values, use the table of values to graph the function, and then identify the function’s domain and range.
$y = 2 \cdot 2^x$
(the table is omitted here as its a chart part, but the relevant x values are -2, -1, 0, 1, 2, 3 with columns for $2^x$ and $y = 2 \cdot 2^x$)

Explanation:

Step1: Calculate \(2^x\) when \(x = -2\)

Recall the rule of negative exponents: \(a^{-n}=\frac{1}{a^n}\). So for \(2^{-2}\), we have \(2^{-2}=\frac{1}{2^2}=\frac{1}{4}\).

Step2: Calculate \(y = 2\cdot2^x\) when \(x = -2\)

Substitute \(2^x=\frac{1}{4}\) into \(y = 2\cdot2^x\). Then \(y = 2\times\frac{1}{4}=\frac{2}{4}=\frac{1}{2}\).

For the other values of \(x\):

• When \(x=-1\):

Step3: Calculate \(2^x\) when \(x = -1\)

Using the negative exponent rule, \(2^{-1}=\frac{1}{2^1}=\frac{1}{2}\).

Step4: Calculate \(y = 2\cdot2^x\) when \(x = -1\)

Substitute \(2^x=\frac{1}{2}\) into \(y = 2\cdot2^x\). Then \(y = 2\times\frac{1}{2}=1\).

• When \(x = 0\):

Step5: Calculate \(2^x\) when \(x = 0\)

Recall that any non - zero number to the power of 0 is 1, so \(2^0 = 1\).

Step6: Calculate \(y = 2\cdot2^x\) when \(x = 0\)

Substitute \(2^x = 1\) into \(y = 2\cdot2^x\). Then \(y=2\times1 = 2\).

• When \(x = 1\):

Step7: Calculate \(2^x\) when \(x = 1\)

\(2^1=2\).

Step8: Calculate \(y = 2\cdot2^x\) when \(x = 1\)

Substitute \(2^x = 2\) into \(y = 2\cdot2^x\). Then \(y = 2\times2=4\).

• When \(x = 2\):

Step9: Calculate \(2^x\) when \(x = 2\)

\(2^2 = 4\).

Step10: Calculate \(y = 2\cdot2^x\) when \(x = 2\)

Substitute \(2^x = 4\) into \(y = 2\cdot2^x\). Then \(y=2\times4 = 8\).

• When \(x = 3\):

Step11: Calculate \(2^x\) when \(x = 3\)

\(2^3=8\).

Step12: Calculate \(y = 2\cdot2^x\) when \(x = 3\)

Substitute \(2^x = 8\) into \(y = 2\cdot2^x\). Then \(y = 2\times8=16\).

Now, for the domain and range:

The function \(y = 2\cdot2^x\) is an exponential function. The domain of an exponential function of the form \(y = a\cdot b^x\) (where \(a>0,b > 0,b
eq1\)) is all real numbers, because we can plug in any real number for \(x\). So the domain is \((-\infty,\infty)\) or \(\{x|x\in\mathbb{R}\}\).

For the range, since \(2^x>0\) for all real \(x\), then multiplying by 2 (a positive number) gives \(2\cdot2^x>0\). So the range is \((0,\infty)\) or \(\{y|y > 0\}\).

Filling the table:

\(x\)	\(2^x\)	\(y = 2\cdot2^x\)
\(-1\)	\(\frac{1}{2}\)	\(1\)
\(0\)	\(1\)	\(2\)
\(1\)	\(2\)	\(4\)
\(2\)	\(4\)	\(8\)
\(3\)	\(8\)	\(16\)

Answer:

• Table filled as above.
• Domain: \(\boldsymbol{(-\infty,\infty)}\) (or all real numbers)
• Range: \(\boldsymbol{(0,\infty)}\) (or all positive real numbers)