Question #80732

1 Answer
Sep 15, 2016

Just stay calm. f'(x) = 3x - 1
Explanation

f(x+h) = 1.5(x+h)^2 - (x+h) + 3.7 = 1.5x^2 + 3xh + 1.5h^2 - x - h + 3.7f(x+h)=1.5(x+h)2(x+h)+3.7=1.5x2+3xh+1.5h2xh+3.7
f(x+h) = f(x) + 3xh + 1.5 h^2 -hf(x+h)=f(x)+3xh+1.5h2h
[f(x+h)-f(x)]/h = 3x - 1 + 1.5hf(x+h)f(x)h=3x1+1.5h

Now take the limit h to 0. The last term drops and you get
f'(x) = lim h->0 [f(x+h) -f(x)]/(h) = 3x -1
Explanation:
f(x+h)=1.5(x+h)^2−(x+h)+3.7=1.5x^2+3xh+1.5h^2−x−h+3.7

[f(x+h)-f(x)]/h = 3x -1 + 1.5h
Taking the limit gives the above result.