How do you use the definition of a derivative to find the derivative of #f(x) = 1.5x^2 - x + 3.7#?

Answer 1

#f'(x)=lim # #h->0 ([1.5(a+h)^2-(a+h)+3.7]-[1.5a^2-a+3.7])/h = (1.5(a^2+2ah+h^2)-a-h+3.7-1.5a^2+a-3.7)/h#
#=(1.5a^2+3ah+1.5h^2-a-h+3.7-1.5a^2+a-3.7)/h=(3ah+1.5h^2-h)/h =3a+1.5h-1=3a-1#

Find f(a+h) and f(a) and then substitute it into the definition and simplify.

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Answer 2

To find the derivative of ( f(x) = 1.5x^2 - x + 3.7 ), you can use the definition of a derivative, which states:

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

Apply this definition to the given function, ( f(x) ), by substituting ( f(x + h) ) and ( f(x) ) into the formula. Then simplify and evaluate the limit as ( h ) approaches zero. This process will give you the derivative of the function, ( f'(x) ).

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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.

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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