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

Question

Emma Aldridge · Accepted Answer

#lim_(Deltax->0) [f(x+Deltax) - f(x)]/(Deltax) #

#f'(x)=lim_(Deltax->0) [2x+Deltax-5]=2x-5# Given: #f(x) = x^2-5x+3# Required: Derivative using limits Solution Strategy: Definition: #(df(x))/(dx)=lim_(Deltax->0) [f(x+Deltax) - f(x)]/(Deltax)# A) Evaluate #f(x)|_(x=x-Deltax)# B) Subtract #f(x)# and simplify A) #f(x)|_(x=x+Deltax)= (x+Deltax)^2-5(x+Deltax)+3# # =cancelx^2+2xDeltax+Deltax^2-cancel(5x)- 5Deltax+cancel3# # - cancelx^2+cancel(5x)-cancel3# #f'(x)=lim_(Deltax->0) [2xcancel(Deltax)+cancel(Deltax^2)^(Deltax)-5cancel(Deltax)]/cancel(Deltax)# #f'(x)=lim_(Deltax->0) [2x+Deltax-5]=2x-5# Good luck!

Charles Arocha · Answer

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

$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

Substitute the function $ f(x) = x^2 - 5x + 3 $ into the formula and simplify to find $ f'(x) $.

Cameron Alexander · Answer

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

$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

Substitute the function $ f(x) = x^2 - 5x + 3 $ into the definition:

$$ f'(x) = \lim_{h \to 0} \frac{(x + h)^2 - 5(x + h) + 3 - (x^2 - 5x + 3)}{h} $$

Expand and simplify:

$$ f'(x) = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - 5x - 5h + 3 - x^2 + 5x - 3}{h} $$

$$ f'(x) = \lim_{h \to 0} \frac{2xh + h^2 - 5h}{h} $$

$$ f'(x) = \lim_{h \to 0} (2x + h - 5) $$

$$ f'(x) = 2x - 5 $$

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