On my homework I am asked to compute the derivative using the defintion of the derivate, and then later am asked to interpret the derivative by plugging in a number. Here is what I had to derive:
h(x)=3x-7 using definition of derivatives I get:
h'(x)=lim (1/h)[g(x+h)-g(x)] = lim (1/h)[3(x+h)-7-(3x-7)]
h->0 h->0
=lim (1/h)[3x+3h-7-3x+7] 3x and -3x cancel, so do 7's
h->0
=lim (1/h)[3h] the h's cancel so I get 3/1
h->0
therefore h'(x)=3 here comes the confusing part
I am asked to compute equation of tangent line to
h(x)=3x-7 at x=-13
I know tangent line eq. is y= f(a)+f ' (a)(x-a)
if we say that f ' (a) is h ' (-13) and a is -13, how in the world do I plug in a -13 into h'(x)=3 ?? Is the answer always 3? I am guessing it is constant since h(x)=3x-7 looks like eq-n of a straight line, but I need confirmation.
2007-09-19
10:42:21
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5 answers
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asked by
pyrojelli
2
in
Mathematics