Which of the following corresponds to the derivative of f(x)=1x+2 at x = 2, using the alternate definition of a derivative, reduced to its simplest form before taking the limit?
The alternate definition of a derivative states that the derivative of a function f(x) at a point x = x0 is given by the limit as h approaches 0 of (f(x0 + h) - f(x0))/h.
Applying this definition to the function f(x) = 1x + 2 at x = 2, we substitute x0 = 2:
(f(2 + h) - f(2))/h
[(1(2 + h) + 2) - (1(2) + 2)]/h
[(2 + h + 2) - (2 + 2)]/h
[(4 + h) - 4]/h
(4 + h - 4)/h
h/h
1
Therefore, the derivative of f(x) = 1x + 2 at x = 2, using the alternate definition, reduced to its simplest form before taking the limit, is 1.
To find the derivative of f(x) = 1x + 2 at x = 2 using the alternate definition of a derivative, we first need to express the alternate definition of the derivative:
f'(x) = lim(h -> 0) [(f(x+h) - f(x)) / h]
Let's substitute the given function f(x) = 1x + 2 into this definition:
f'(x) = lim(h -> 0) [(f(x+h) - f(x)) / h]
= lim(h -> 0) [((1(x + h) + 2) - (1x + 2)) / h]
= lim(h -> 0) [(x + h + 2 - x - 2) / h]
Now, simplify the expression within the limit:
f'(x) = lim(h -> 0) [(h) / h]
The h cancels out:
f'(x) = lim(h -> 0) [1]
Therefore, the derivative of f(x) = 1x + 2 at x = 2, using the alternate definition of a derivative, is equal to 1.
To find the derivative of a function using the alternate definition, we can use the formula:
\(f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h}\)
In this case, we need to find the derivative of \(f(x) = 1x + 2\) at \(x = 2\). Let's substitute these values into the formula:
\(f'(2) = \lim_{{h \to 0}} \frac{{f(2+h) - f(2)}}{h}\)
Now, we can calculate the values of \(f(2+h)\) and \(f(2)\):
\(f(2+h) = 1(2+h) + 2 = 2 + h + 2 = h + 4\)
\(f(2) = 1(2) + 2 = 2 + 2 = 4\)
Substituting these values into the formula, we get:
\(f'(2) = \lim_{{h \to 0}} \frac{{(h + 4) - 4}}{h}\)
Simplifying further:
\(f'(2) = \lim_{{h \to 0}} \frac{{h}}{h}\)
As \(h\) approaches 0, the numerator and denominator both approach 0. Therefore, the limit of this expression is 1:
\(f'(2) = 1\)
So, the derivative of \(f(x) = 1x + 2\) at \(x = 2\) using the alternate definition is 1.