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

The answer is Choice B.