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Math Help - Tangent to the curve

  1. #1
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    Tangent to the curve

    I need help finding the tangent to the curve y = \frac{3}{1 + \sqrt{x}}, at the point (4,1). Im really only having problems finding the gradent of the tangent, i already know the answer but everytime i try i either get the gradent as 0 or infinity which isnt correct.

    Im not ment to be using differentiation, but instead using limits.
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    Hello,
    Quote Originally Posted by Lipticboven View Post
    I need help finding the tangent to the curve y = \frac{3}{1 + \sqrt{x}}, at the point (4,1). Im really only having problems finding the gradent of the tangent, i already know the answer but everytime i try i either get the gradent as 0 or infinity which isnt correct.

    Im not ment to be using differentiation, but instead using limits.
    Let f(x)=\frac{3}{1+\sqrt{x}}

    You have to find f'(4) in order to find the gradient.

    So since you have to use limits and not differentiation, remember that :
    f'(a)=\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}

    So here :
    f'(4)=\lim_{h \to 0} \frac{f(4+h)-f(4)}{h}

    =\lim_{h \to 0} \frac{\frac{3}{1+\sqrt{4+h}}-\frac{3}{1+\sqrt{4}}}{h}

    =\lim_{h \to 0} \frac{\frac{3}{1+\sqrt{4+h}}-1}{h} \cdot \frac{1+\sqrt{4+h}}{1+\sqrt{4+h}}

    =\lim_{h \to 0} \frac{3-(1+\sqrt{4+h})}{h(1+\sqrt{4+h})}

    =\lim_{h \to 0} \frac{2-\sqrt{4+h}}{h(1+\sqrt{4+h})} \cdot \frac{2+\sqrt{4+h}}{2+\sqrt{4+h}}

    (it's multiplying by the conjugate of the numerator and then use the identity (a-b)(a+b)=a^2-b^2)

    =\lim_{h \to 0} \frac{4-(4+h)}{h(1+\sqrt{4+h})(2+\sqrt{4+h})}

    =\lim_{h \to 0} \frac{-h}{h(1+\sqrt{4+h})(2+\sqrt{4+h})}

    =\lim_{h \to 0} \frac{-1}{(1+\sqrt{4+h})(2+\sqrt{4+h})}


    And now the limit is defined...
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