Question

For j(x) = 3x − 1, find j of the quantity x plus h end quantity minus j of x all over h period3 to the power of the quantity x minus 1 end quantity times the quantity 3 to the power of h end quantity all over h3 to the power of the quantity x minus 1 end quantity times the quantity 3 to the power of h minus 1 end quantity all over h3 to the power of the quantity x minus 1 end quantity times the quantity 3 to the power of h plus 1 end quantity all over hthe quantity x minus 1 end quantity times the quantity 3 to the power of h plus 1 end quantity all over h

Answer 1

From the problem, we have :

$j(x)=3^{x-1}$

j(x + h) will be :

$\begin{gathered} j(x+h)=3^{x+h-1} \\ =3^h(3^{x-1}) \end{gathered}$

Note that in multiplying expressions with exponents :

$x^a(x^b)=x^{a+b}$

The exponent were added in the result, doing in reverse with the given expression :

$\begin{gathered} 3^{x+h-1}=3^{h+(x-1)} \\ =3^h(3^{x-1}) \end{gathered}$

Evaluating the required expression :

$\begin{gathered} \frac{j(x+h)-j(x)}{h}=\frac{3^h(3^{x-1})-3^{x-1}}{h} \\ =\frac{3^{x-1}(3^h-1)}{h} \end{gathered}$

The answer is Choice B.