Help Finding Derivative and its Interpretation?

Question

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.

let me put the question in its original form so you can see what I mean.

For this problem use the definition of the derivative to compute the derivative of the given function.

g(x)=3x-7

then use the derivative found in the previous part to answer this question.

Find the equation of the tangent line to
h(x)=3x-7 at x=-13 PLEASE HELP

Answer 1

Kay, so you found the derivative.

Plug in the x value into the derivative and find slope of tangent line.

h'(x) = 3

y = 3x + b

Find a point to put in.

Remember that it touches x = -13 and y=h(x) = ??

Find that h(x).

h(-13) = -39 - 7

h(-13) = -46

-46 = 3(-13) + b

-46 = -39 + b

b = 7

y = 3x + 7

And you can know this is true because a line tangent to another STRAIGHT line is the line itself.

Answer 2

The derivative tells you the SLOPE of the tangent line. h'(x) is the constant 3, so the slope of the tangent is always 3, regardless of x. But remember that the line and the original curve share a point at the tangent point. Sin h(-13) = -46, we're looking for the line whose slope is 3 and passes through (-13, -46).

Answer 3

the original equation h(x) is the equation of a line. For a line to be tangent to another line they must be the same line. therefore since you get an inconsistent answer, there is no equation of a line tangent to h(x)=3x-7 at x=-13. at x=-13 h(x) = -46. That is the only equation that satisfies the x=-13 point. it is a point and not a line. The derivative of a line is a point. Since h'(x) =3 then the point in question is 3,2 which is inconsistent with x=-13. Your conclusions are correct.

Answer 4

y = 3x-7 is a straight line with slope = 3 and y-intercept = -7.

A line does not have a tangent. Only a curve can have a tangent. A tangent is a line that touches a curve at only one place such as a tangent to a circle. Thus a line can not have a tangent.

Are you sure you are stating the problem correctly?

Answer 5

yes, h'(-13)=3, because h'(x)=3 for any x real.