Question 639699: If g(x)=x^2-2x+5, find and simplify g(x+h)-g(x)/h
g(x+h) = (x+h)^2 - 2(x+h) + 5 = x^2 + 2xh + h^2 -2x - 2h + 5
Then, g(x+h)-g(x) = (x^2 + 2xh + h^2 -2x - 2h + 5) - (x^2-2x+5)
= (2xh + h^2 - 2h)
Finally, g(x+h)-g(x)/h = (2xh + h^2 - 2h)/h. From the numerator we factor out an 'h'. Numerator = (2xh + h^2 - 2h) = h*(2x + h -2). Therefore g(x+h)-g(x)/h = (h*(2x + h -2)) / h = (2x + h - 2). Now generally when taking this kind of limit for a derivative the value of h 'approaches' 0. So now when we substitute h for 0 we see the derivative (lim of g(x+h)-g(x)/h as h-> 0) approaches (2x - 2) which is the desired derivative!
Answer by tinbar(133)   (Show Source):
You can put this solution on YOUR website! If g(x)=x^2-2x+5, find and simplify g(x+h)-g(x)/h
g(x+h) = (x+h)^2 - 2(x+h) + 5 = x^2 + 2xh + h^2 -2x - 2h + 5
Then, g(x+h)-g(x) = (x^2 + 2xh + h^2 -2x - 2h + 5) - (x^2-2x+5)
= (2xh + h^2 - 2h)
Finally, g(x+h)-g(x)/h = (2xh + h^2 - 2h)/h. From the numerator we factor out an 'h'. Numerator = (2xh + h^2 - 2h) = h*(2x + h -2). Therefore g(x+h)-g(x)/h = (h*(2x + h -2)) / h = (2x + h - 2). Now generally when taking this kind of limit for a derivative the value of h 'approaches' 0. So now when we substitute h for 0 we see the derivative (lim of g(x+h)-g(x)/h as h-> 0) approaches (2x - 2) which is the desired derivative!
|
|
|
