Using the limit definition, how do you find the derivative of #f(x) = -5x^2+8x+2#?

Question

Charles Arocha · Accepted Answer

#f'(x)= -10x+8# Using the limit definition to find the derivativeof #f(x)# #f'(x)=lim_(h->0)(f(x+h)-f(x))/h# so it can be expressed as following #lim_(h->0)((-5(x+h)^2+8(x+h)+2)-(-5x^2+8x+2))/h# =#-10x+8#

Aurora Alvarado · Answer

Use the formula #f'(x) = lim_(h-> 0) (f(x + h) - f(x))/h#

#f'(x) = lim_(h->0) (-5(x + h)^2 + 8(x + h) + 2 -(-5x^2 + 8x + 2))/h#

#f'(x) = lim_(h->0)(-5(x^2 + 2xh + h^2) + 8x + 8h + 2 + 5x^2 - 8x - 2)/h#

#f'(x)= lim_(h->0) (-5x^2 - 10xh - 5h^2 + 8x + 8h + 2 + 5x^2 - 8x - 2)/h#

#f'(x) = lim_(h->0) (-10xh - 5h^2+ 8h)/h#

#f'(x) = lim_(h->0) (cancel(h)(-10x - 5h + 8))/cancel(h)#

#f'(x) = -10x - 5(0) + 8#

#f'(x) = -10x + 8#

#:.# The derivative is #dy/dx = -10x + 8#.

Hopefully this helps!

William Abdalla · Answer

undefined To find the derivative of $ f(x) = -5x^2 + 8x + 2 $ using the limit definition, we apply the formula:

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

First, we substitute the given function into the formula:

$$ f'(x) = \lim_{h \to 0} \frac{(-5(x + h)^2 + 8(x + h) + 2) - (-5x^2 + 8x + 2)}{h} $$

Next, we expand and simplify the expression:

$$ f'(x) = \lim_{h \to 0} \frac{-5(x^2 + 2xh + h^2) + 8x + 8h + 2 + 5x^2 - 8x - 2}{h} $$
$$ f'(x) = \lim_{h \to 0} \frac{-5x^2 - 10xh - 5h^2 + 8x + 8h + 2 + 5x^2 - 8x - 2}{h} $$
$$ f'(x) = \lim_{h \to 0} \frac{-10xh - 5h^2 + 8h}{h} $$

Now, we cancel out common terms and simplify further:

$$ f'(x) = \lim_{h \to 0} -10x - 5h + 8 $$
$$ f'(x) = -10x + 8 $$

Therefore, the derivative of $ f(x) = -5x^2 + 8x + 2 $ is $ f'(x) = -10x + 8 $.