f(x) = x2 + 2x - 1
Compute the derivative (that is, the derivative function) by evaluating the limit. Show your work. (Notice that for the function k(x),
you’ll have to compute the limit differently depending on whether x < 2 or x > 2, and
you’ll end up with a piecewise function for the derivative. Since the graph of k(x) is
made of straight lines, you can also find its derivative by looking at the actual slopes.)
Typical solution for this kind of problem
f(x) = x^2 + 2x + 1
f(x+h) = (x+h)^2 + 2(x+h) + 1
= x^2 + 2hx + h^2 + 2x + 2h + 1
d(f(x))/dx = lim [f(x+h) - f(x)]/h , as h ---> 0
= lim [x^2 + 2hx + h^2 + 2x + 2h + 1 - (x^2 + 2x + 1)]/(x+h - x) , h ---> 0
= lim (x^2 + 2hx + h^2 + 2x + 2h + 1 - x^2 - 2x - 1)/h , as h ---> 0
= lim (2hx + h^2 + 2h)/h as h --->0
= lim 2x + h + 2 , as h --->0
= 2x + 2
no idea where k(x) comes into play
To compute the derivative of the function f(x) = x^2 + 2x - 1, we'll use the limit definition of the derivative:
f'(x) = lim(h->0) [f(x+h) - f(x)] / h
Let's evaluate this limit step by step:
Step 1: Calculate f(x+h)
f(x + h) = (x + h)^2 + 2(x + h) - 1
= x^2 + 2hx + h^2 + 2x + 2h - 1
Step 2: Calculate f(x + h) - f(x)
f(x + h) - f(x) = (x^2 + 2hx + h^2 + 2x + 2h - 1) - (x^2 + 2x - 1)
= 2hx + h^2 + 2h
Step 3: Divide by h
[f(x + h) - f(x)] / h = (2hx + h^2 + 2h) / h
= 2x + h + 2
Step 4: Take the limit as h approaches 0
lim(h->0) [2x + h + 2] = 2x + 2
Thus, the derivative function of f(x) = x^2 + 2x - 1 is f'(x) = 2x + 2.
However, for the function k(x), we need to consider x < 2 and x > 2 separately.
For x < 2:
The limit definition of the derivative remains the same:
k'(x) = lim(h->0) [k(x+h) - k(x)] / h
Step 1: Calculate k(x+h)
k(x + h) = (x + h)^2 + 2(x + h) - 1
= x^2 + 2hx + h^2 + 2x + 2h - 1
Step 2: Calculate k(x + h) - k(x)
k(x + h) - k(x) = (x^2 + 2hx + h^2 + 2x + 2h - 1) - (x^2 + 2x - 1)
= 2hx + h^2 + 2h
Step 3: Divide by h
[k(x + h) - k(x)] / h = (2hx + h^2 + 2h) / h
= 2x + h + 2
Step 4: Take the limit as h approaches 0
lim(h->0) [2x + h + 2] = 2x + 2
For x > 2:
We'll calculate the slope of the function directly, as it consists of straight lines.
k(x) = x^2 + 2x - 1, for x > 2
The derivative of k(x) will be the slope of the line connecting x = 2 to any point greater than 2 on the graph of k(x).
The slope of this line is equal to the derivative at x = 2, which we found to be 2x + 2.
Therefore, for x > 2, the derivative of k(x) is also equal to 2x + 2.
In conclusion, the derivative (derivative function) of k(x) = x^2 + 2x - 1 is given by:
k'(x) = 2x + 2 for x < 2
2x + 2 for x > 2
To compute the derivative of the function f(x) = x^2 + 2x - 1 using the limit definition of a derivative, we follow these steps:
Step 1: Write down the limit definition of a derivative:
The derivative of f(x) is given by the following limit:
f'(x) = lim(h → 0) [f(x + h) - f(x)] / h
Step 2: Substitute the function f(x) into the limit definition:
f'(x) = lim(h → 0) [(x + h)^2 + 2(x + h) - 1 - (x^2 + 2x - 1)] / h
Step 3: Simplify the expression inside the limit:
f'(x) = lim(h → 0) [x^2 + 2xh + h^2 + 2x + 2h - 1 - x^2 - 2x + 1] / h
Step 4: Combine like terms and simplify further:
f'(x) = lim(h → 0) [2xh + h^2 + 2h] / h
= lim(h → 0) [h(2x + h + 2)] / h
Step 5: Cancel out a common factor of h:
f'(x) = lim(h → 0) 2x + h + 2
Step 6: Evaluate the limit as h approaches 0:
In this case, we do not need to evaluate the limit separately for x < 2 and x > 2, as the expression is continuous everywhere.
So, the derivative function, f'(x), will be:
f'(x) = 2x + 2
Therefore, the derivative function of f(x) = x^2 + 2x - 1 is f'(x) = 2x + 2.