Posted by T.GRacken on February 22, 2001 at 15:47:30:
In Reply to: Please help me check this problem posted by Ben on February 21, 2001 at 23:16:40:
: 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
>>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)
>>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.