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

Question

Emma Aldridge · Accepted Answer

#f'(x)=6x-5#

The limit definition of a derivative states that

#f'(x)=lim_(hrarr0)(f(x+h)-f(x))/h#

Substituting #f(x)=3x^2-5x+2# into #f'(x)#,

#f'(x)=lim_(hrarr0)((3(x+h)^2-5(x+h)+2)-(3x^2-5x+2))/h#

From this point on, you want to expand and simplify.

#f'(x)=lim_(hrarr0)(3x^2+6xh+3h^2-5x-5h+2-3x^2+5x-2)/h#

#f'(x)=lim_(hrarr0)(color(red)cancelcolor(black)(3x^2)+6xh+3h^2color(blue)cancelcolor(black)(-5x)-5hcolor(darkorange)cancelcolor(black)(+2)color(red)cancelcolor(black)(-3x^2)color(blue)cancelcolor(black)(+5x)color(darkorange)cancelcolor(black)(-2))/h#

#f'(x)=lim_(hrarr0)(6xh+3h^2-5h)/h#

Factor out #h# from the numerator.

#f'(x)=lim_(hrarr0)(h(6x+3h-5))/h#

#f'(x)=lim_(hrarr0)6x+3h-5#

Plugging in #h=0#,

#f'(x)=6x+3(0)-5#

#color(green)(|bar(ul(color(white)(a/a)color(black)(f'(x)=6x-5)color(white)(a/a)|)))#

Samantha Adkison · Answer

undefined To find $ f'(x) $ using the limit definition, you apply the definition of the derivative:

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

For the function $ f(x) = 3x^2 - 5x + 2 $, you substitute into the formula:

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

Then, simplify and evaluate the limit.