Finding points on a tangent line

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To find points on the graph of g(x) = (1/3)x^3 - (1/2)x^2 - 4x + 8 where the tangent line has a slope of -4, the derivative g'(x) was calculated as x^2 - x - 4. Setting this equal to -4 led to the equation x^2 - x = 0, yielding potential solutions x = 0 and x = -1. However, a mistake was identified in solving the quadratic, as x = -1 is not a valid solution. The correct y-values were computed for x = 0, resulting in the point (0, 8). The discussion highlights the importance of careful algebraic manipulation in finding tangent points.
Vandella
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Homework Statement



Find the point(s) on the graph of the function at which the tangent line has the indicated slope. (If an answer does not exist, enter DNE.)

g(x) = (1/3)x^3 - (1/2)x^2 - 4x +8

mtan=-4

Homework Equations





3. The Attempt at a Solution

firstly i derived g(x) to give x^2-x-4
as tangent line = -4 substituted into equation to give x^2-x-4=-4
manipulated to get x^2-x=0
that gave me x=0 or x=-1
used those values in original function to find y values
points i obtained were (0,8) and (-1,67/6)

when i enter these with lowest x value first it says i am wrong please help
 
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Since this is just a silly brainfart mistake I'm going to give you the answer, x=-1 is not a solution, you made a silly mistake when you solved the quadratic.
 
ahhh i can't believe i missed that

thanks
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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