PLS HELP ILL MARK YOU BRAINEST DUE SOON
2 answers:
You might be interested in
Given a function <em>g(x)</em>, its derivative, if it exists, is equal to the limit
The limit is some expression that is itself a function of <em>x</em>. Then the derivative of <em>g(x)</em> at <em>x</em> = 1 is obtained by just plugging <em>x</em> = 1. In other words, find <em>g'(x)</em> - and this can be done with or without taking a limit - then evaluate <em>g'</em> (1).
Alternatively, you can directly find the derivative at a point by computing the limit
But this is essentially the same as the first method, we're just replacing <em>x</em> with 1.
Yet another way is to compute the limit
but this is really the same limit with <em>h</em> = <em>x</em> - 1.
You do not compute <em>g</em> (1) first, because as you say, that's just a constant, so its derivative is zero. But you're not concerned with the derivative of some <em>number</em>, you care about the derivative of a function that depends on a <em>variable.</em>
Answer:
The answer to your question is f(x) =-
Step-by-step explanation:
g(x) = 9x - 8
f(x) = ?
h(x) = (f°g)(x) = 9 / (9x - 8)
Process
When we evaluate a function with another function, just substitute the second function in all the x of the first function.
From the information given we notice that g(x) = 9x - 8 and it is in the denominator of the function h(x). This means that before we substitute this value in f(x) there, there was a "x".
Then
f(x) =- so (f°g)(x) =-
or h(x) = -