How do you use the limit definition to find the derivative of #y=-3x^2+x+4#?

Question

Isaiah Allen · Accepted Answer

-6x+1

Using limit definition derivative of y, that is f(x) would be #f'(x)= Lim_(h->0) (f(x+h) -f(x))/h# = #Lim_(h->0) (-3(x+h)^2 +x+h +4 - (-3x^2 +x+4))/h# =#Lim_(h->0) (-3x^2 -6xh -3h^2 +x+h+4 +3x^2 -x-4)/h# = #Lim_(h->0) (-6xh-3h^2 +h)/h# =#Lim_(h->0) -6x-3h+1# =-6x+1

Evelyn Agard · Answer

undefined To find the derivative of $ y = -3x^2 + x + 4 $ using the limit definition, 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 $ f(x) = -3x^2 + x + 4 $ into the formula.

3. Simplify the expression:

$$ f'(x) = \lim_{h \to 0} \frac{-3(x + h)^2 + (x + h) + 4 - (-3x^2 + x + 4)}{h} $$

4. Expand and simplify the expression inside the limit:

$$ f'(x) = \lim_{h \to 0} \frac{-3(x^2 + 2hx + h^2) + x + h + 4 + 3x^2 - x - 4}{h} $$

$$ f'(x) = \lim_{h \to 0} \frac{-3x^2 - 6hx - 3h^2 + x + h + 4 + 3x^2 - x - 4}{h} $$

$$ f'(x) = \lim_{h \to 0} \frac{-6hx - 3h^2 + h}{h} $$

5. Factor out $ h $ from the numerator:

$$ f'(x) = \lim_{h \to 0} \frac{h(-6x - 3h + 1)}{h} $$

6. Cancel out $ h $ from the numerator and denominator:

$$ f'(x) = \lim_{h \to 0} -6x - 3h + 1 $$

7. Evaluate the limit as $ h $ approaches 0:

$$ f'(x) = -6x + 1 $$

So, the derivative of $ y = -3x^2 + x + 4 $ is $ f'(x) = -6x + 1 $.