How do you use the limit definition of the derivative to find the derivative of #f(x)=-3x^2+x+5#?

Question

Evelyn Agard · Accepted Answer

# f'(x)=-6x+1 # By definition of the derivative # f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h # So with # f(x) = -3x^2+x+5 # we have; # f'(x)=lim_(h rarr 0) ( {-3(x+h)^2+(x+h)+5 } - {-3x^2+x+5 } ) / h # # :. f'(x)=lim_(h rarr 0) ( {-3(x^2+2hx+h^2)+x+h+5 } - {-3x^2+x+5 } ) / h # # :. f'(x)=lim_(h rarr 0) ( -3x^2-6hx-3h^2+x+h+5 +3x^2-x-5 ) / h # # :. f'(x)=lim_(h rarr 0) ( -6hx-2h^2+h ) / h # # :. f'(x)=lim_(h rarr 0) ( -6x-2h+1 ) # # :. f'(x)=-6x+1 #

Christopher Agresta · Answer

undefined To find the derivative of $ f(x) = -3x^2 + x + 5 $ using the limit definition of the derivative, follow these steps:

1. Write down the limit definition of the derivative:
$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

2. Substitute the function $ f(x) = -3x^2 + x + 5 $ into the limit definition:
$$ f'(x) = \lim_{h \to 0} \frac{[-3(x + h)^2 + (x + h) + 5] - [-3x^2 + x + 5]}{h} $$

3. Expand and simplify the expression inside the limit:
$$ f'(x) = \lim_{h \to 0} \frac{-3(x^2 + 2xh + h^2) + x + h + 5 + 3x^2 - x - 5}{h} $$
$$ f'(x) = \lim_{h \to 0} \frac{-3x^2 - 6xh - 3h^2 + x + h + 5 + 3x^2 - x - 5}{h} $$
$$ f'(x) = \lim_{h \to 0} \frac{-6xh - 3h^2 + h}{h} $$

4. Factor out $ h $ and cancel:
$$ f'(x) = \lim_{h \to 0} \frac{h(-6x - 3h + 1)}{h} $$
$$ f'(x) = \lim_{h \to 0} (-6x - 3h + 1) $$

5. Evaluate the limit as $ h $ approaches 0:
$$ f'(x) = -6x + 1 $$

Therefore, the derivative of $ f(x) = -3x^2 + x + 5 $ is $ f'(x) = -6x + 1 $.