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How do you use the limit definition of the derivative to find the derivative of #f(x)=2x+11#?

Answer 1

# f'(x)= 2 #

By definition, # f'(x)= lim_(h->0)(f(x+h)-f(x))/h#

so for #f(x)=2x+11# we have: # f'(x)= lim_(h->0)({2(x+h)+11 - (2x+11)})/h# # f'(x)= lim_(h->0)( 2x+2h+11 - 2x-11 )/h# # f'(x)= lim_(h->0)( 2h )/h# # f'(x)= lim_(h->0)( 2 )# # f'(x)= 2 #

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Answer 2

Ezekiel Alexander

To use the limit definition of the derivative to find the derivative of ( f(x) = 2x + 11 ), we start by applying the definition:

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

Substitute the function ( f(x) = 2x + 11 ) into the formula:

[ f'(x) = \lim_{h \to 0} \frac{(2(x + h) + 11) - (2x + 11)}{h} ]

Simplify the expression inside the limit:

[ f'(x) = \lim_{h \to 0} \frac{2x + 2h + 11 - 2x - 11}{h} ]

Combine like terms:

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

Cancel out the common factor of ( h ):

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

Since the limit of a constant is the constant itself, we get:

[ f'(x) = 2 ]

Therefore, the derivative of ( f(x) = 2x + 11 ) using the limit definition is ( f'(x) = 2 ).

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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.