Question

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

Answer 1

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
\[f'(x) = -1/(x+1)(x+0+1) = -1/(x+1)^{2}\]

Answer 2

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

Answer 3

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

Answer 4

thnx a lot ,,,,,,,,,,,,,,,,,,,,,,,,