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

Answer 1

Christopher Allers

#f'(x)=4x-1#

The definition of the derivative is

#lim_(h->0) (f(x+h)-f(x))/h#

Proceeding as follows

#lim_(h->0)[[2(x+h)^2-(x+h)+5]-[2x^2-x+5]]/h#

#lim_(h->0)[[2(x^2+2xh+h^2)-x-h+5]-[2x^2-x+5]]/h#

#lim_(h->0)(2x^2+4xh+2h^2-x-h+5-2x^2+x-5)/h#

#lim_(h->0)(4xh+2h^2-h)/h#

#lim_(h->0)(h(4x+2h-1))/h#

#lim_(h->0)4x+2h-1=4x+2(0)-1=4x-1#

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Answer 2

Christopher Agresta

To find ( f'(x) ) using the definition of a derivative for ( f(x) = 2x^2 - x + 5 ), we use the formula:

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

Substitute ( f(x) = 2x^2 - x + 5 ) into the formula:

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

Expand ( (2(x+h)^2 - (x+h) + 5) ):

[ f'(x) = \lim_{h \to 0} \frac{(2x^2 + 4xh + 2h^2 - x - h + 5) - (2x^2 - x + 5)}{h} ]

[ f'(x) = \lim_{h \to 0} \frac{2x^2 + 4xh + 2h^2 - x - h + 5 - 2x^2 + x - 5}{h} ]

[ f'(x) = \lim_{h \to 0} \frac{4xh + 2h^2 - h}{h} ]

[ f'(x) = \lim_{h \to 0} \frac{h(4x + 2h - 1)}{h} ]

[ f'(x) = \lim_{h \to 0} (4x + 2h - 1) ]

[ f'(x) = 4x - 1 ]

So, ( f'(x) = 4x - 1 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.