Question

[HELP] For j(x) = le" class="latex-formula">, find j of the quantity x plus h end quantity minus j of x all over h period
the quantity x minus 3 end quantity times the quantity 5 to the power of h plus 1 end quantity all over h $\frac{(x-3)(5^2+1)}{h}$

5 to the power of the quantity x minus 3 end quantity times the quantity 5 to the power of h end quantity all over h $\frac{5^x^-^3(h^h)}{h}$

5 to the power of the quantity x minus 3 end quantity times the quantity 5 to the power of h plus 1 end quantity all over h $\frac{5^x^-^3(5^h+1)}{h}$

5 to the power of the quantity x minus 3 end quantity times the quantity 5 to the power of h minus 1 end quantity all over h $\frac{5^x^-^3(5^h-1)}{h}$

Answer 1

The difference quotient of the function that has been presented to us will turn out to be 5.

How can I calculate the quotient of differences?

In this step, we wish to determine the difference quotient for the function that was supplied.

To begin, keep in mind that the difference quotient may be calculated by:

Lim h->0 $\frac{f(x+h)-f(x)}{h}$

Now, for the purpose of the function, we need this:

Then we will have:

$\begin{aligned}&\lim _{h \rightarrow 0} \frac{j(x+h)-j(x)}{h} \\&\lim _{h \rightarrow 0} \frac{5 *(x+h)-3-5 * x+3}{h} \\&\lim _{h \rightarrow 0} \frac{5 x+5 h-3-5 x+3}{h} \\&\lim _{h \rightarrow 0} \frac{5 h}{h}=5\end{aligned}$

j(x) = 5x - 3

Then the following will be true:

Therefore, 5 is the value of the difference quotient for j(x) is %

Read the following if you are interested in finding out more about difference quotients:

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