Posted by T.G. on February 22, 2001 at 16:33:14:
In Reply to: Big Oops! posted by T.GRacken on February 22, 2001 at 15:47:30:
: : Find the slope and the equation of the tangent line to the graph of the function at the given value of x.
: :
: : y=-x^-3 + 5x^-1 + x ;x=2
: : I have derived and worked the problem and got this answer. m= 14.25
: : y=14.25x + .154
: : is the a correct answer?
: : Thanks.
:
: Sorry! I made a boo boo. (however, your answer is still incorrect)
: I wrote:
: >>Determine y' (i.e. take the derivative... careful with the negative exponents!)
: I should have been careful myself. I missed the negative coefficient on x^-3.
: >>Evaluate y' when x=2 (I get -7/16),
: SHOULD BE 35/8
the 35/8 is y when x is 2!
:
: >>and this will be the slope of the line tangent to the graph of y when x is 2.
: >>determine the value of y (not y') when x=2. This will give you the other coordinate for the point on the graph when x=2...
: >>You should get y = 37/8 when x=2.
: NO! YOU SHOULD GET -1/16.
: AND THE POINT SHOULD BE (2, 35/8)
the -1/16 is y' when x is 2. the point is o.k.
I just wrote the value for y and y' in the wrong place.
: >>Thus, the point (2, 37/8) is the point on the graph that the tangent line will pass through.
: >>Use the point/slope formula for determining a line (or if you prefer a different method that is correct, use it)
: >>You should end up with y=(-7/16)x + (11/2)
: AGAIN, NO! YOU SHOULD END UP WITH y=(-1/16)x + (9/2).
:
: Sorry. And it's not screaming, the caps are just to make my errors standout.
: Of course, if the original problem were y=x^-3 + 5x^-1 + x then I stand by my earlier calculations.