How do you use the definition of a derivative to find the derivative of # f(x) = 5x + 9# at x=2?
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To find the derivative of the function ( f(x) = 5x + 9 ) at ( x = 2 ) using the definition of a derivative:
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Start with the definition of the derivative of a function ( f(x) ) at a point ( x = a ): [ f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} ]
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Substitute ( f(x) = 5x + 9 ) into the definition: [ f'(2) = \lim_{h \to 0} \frac{f(2 + h) - f(2)}{h} ]
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Evaluate ( f(2 + h) ) and ( f(2) ): [ f(2 + h) = 5(2 + h) + 9 = 10 + 5h + 9 = 19 + 5h ] [ f(2) = 5(2) + 9 = 10 + 9 = 19 ]
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Substitute the expressions for ( f(2 + h) ) and ( f(2) ) into the derivative definition: [ f'(2) = \lim_{h \to 0} \frac{(19 + 5h) - 19}{h} ]
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Simplify the expression: [ f'(2) = \lim_{h \to 0} \frac{5h}{h} ] [ f'(2) = \lim_{h \to 0} 5 ]
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Evaluate the limit: [ f'(2) = 5 ]
So, the derivative of ( f(x) = 5x + 9 ) at ( x = 2 ) is ( f'(2) = 5 ).
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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.
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.
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.
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.
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