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anonymous

  • 5 years ago

h(x) = 1/x+1 use the definitionto find the derivative of h at x = 0 and then the tangent line to the graph of h at this point ? plz help i have exam tomarow

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  1. dumbcow
    • 5 years ago
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    definition of the derivative is \[f'(x) = \lim_{h \rightarrow 0} (f(x+h) - f(x))/h\] substitute x+h in for x in the function \[[(1/(x+h+1) - (1/(x+1))]/h\] combine fractions using common denominator (x+1)(x+h+1) and add like terms \[-h/[h(x+1)(x+h+1)]\] Notice how h will now cancel on top and bottom of fraction. Next evaluate the limit at h = 0 so plug in 0 for h \[f'(x) = -1/(x+1)(x+0+1) = -1/(x+1)^{2}\]

  2. anonymous
    • 5 years ago
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    \[\int\limits \sum_{n=1}^{\infty}\frac{5+x}{u^2 -6} du\]

  3. dumbcow
    • 5 years ago
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    To evaluate the derivative at a point plug in the desired x-value For x = 0 f'(0) = -1/(0+1)^2 = -1 Remember the derivative gives the slope at that point. So at x =0 the slope is -1 A tangent line is a linear line of the form y = mx+b The key to this problem is to solve for b your y-intercept plug in known values y = f(0) = 1/(0+1) = 1 m = -1 x = 0 -> 1 = -1*(0) + b -> b = 1 Finally our tangent line at x=0 is y = -x +1 Hope this helps, good luck on the exam

  4. anonymous
    • 5 years ago
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    thnx a lot ,,,,,,,,,,,,,,,,,,,,,,,,

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