Use the limit definition of the derivative to find the instantaneous rate of change of f(x)=5x^2+3x+3 at x=4

Question

Use the limit definition of the derivative to find the instantaneous rate of change of
f(x)=5x^2+3x+3 at x=4

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MichaelMet 1 month 2021-08-15T21:06:45+00:00 1 Answers 0 views 0

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    2021-08-15T21:07:49+00:00

    f'(4) = 43

    Explanation:

    Given: f(x)=5x^2 + 3x + 3

    \displaystyle f'(x)= \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}

    Note that

    f(x+h) = 5(x+h)^2 + 3(x+h) + 3

    \:\:\;\:\:\:\:= 5(x^2 + 2hx + h^2) + 3x + 3h +3

    \:\:\;\:\:\:\:= 5x^2 + 10hx + 5h^2 + 3x + 3h +3

    Substituting the above equation into the expression for f'(x), we can then write f'(x) as

    \displaystyle f'(x) = \lim_{h \to 0} \dfrac{10hx + 3h + 5h^2}{h}

    \displaystyle\:\:\;\:\:\:\:= \lim_{h \to 0} (10x +3 +5h)

    \:\:\;\:\:\:\:= 10x + 3

    Therefore,

    f'(4) = 10(4) + 3 = 43

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