when the (x,y) coordinates are indicated at an extremum?
Assume the curve is concave down and it has a maximum at
(-3, -6), is it possible to find the value of the derivative
given the graph and the (x,y) coordinates
?
2006-08-19 08:53:24 · 9 answers · asked by tjhauck2001 2 in Science & Mathematics ➔ Mathematics
If the point is located at an extremum, the derivative is zero.
2006-08-19 08:59:46 · answer #1 · answered by Pascal 7 · 0⤊ 0⤋
At a relative extreme point, the derivative is always zero, because the tangent line is horizontal. So for your specific example, the value of the derivative at (-3, -6) is zero. But given a different graph, or a different point that is not a maximum or minimum, it would be easier if you had the equation of the curve.
2006-08-19 17:00:54 · answer #2 · answered by Animate M 1 · 0⤊ 0⤋
You can find the derivative (the volume generated rotating an irregular shaped item around a 360 degree axis) this is done by finding the values for X and Y as X approaches infinity and Y approaches 0 in the equation you have given.
2006-08-19 16:19:44 · answer #3 · answered by Anonymous · 0⤊ 0⤋
draw a tangent to the graph at any point , the slope of the tangent is approximately the derivative , you can find the slop by measuring the angle between the tangent and the x axis and then computing the tan of the angle ( use a calculator directly )
, on the maximum the tangent is horizontal the the slope is 0 the tan(0) = 0
2006-08-19 16:12:39 · answer #4 · answered by khaled 2 · 0⤊ 0⤋
Draw a tangent to the curve at the point where you want to know the value of the derivative. Measure the slope of that tangent line and that's the numerical value of the derivative at that point.
BTW. The slope at a relative maxima or minima is zero.
Doug
2006-08-19 16:07:50 · answer #5 · answered by doug_donaghue 7 · 0⤊ 0⤋
Draw a tangent at the given point (-3,-6).Drop perpendiculars on the tangent along the x and y axis.U will get a triangle.Find out the value of dy/dx, of the triangle.That will be the derivative value at that point.
2006-08-19 16:02:37 · answer #6 · answered by Vin 1 · 0⤊ 0⤋
Assuming that this curve is quadratic:
y = -(x + 3)^2 - 6
Since this is y = X^2 inverted (to get the -)
and the vertex (max/min) has been transformed from (0,0) to (-3, -6)
Multiply out your equation to get
y = - (x^2 + 6x + 9) - 6
=>
y = -x^2 - 6x - 15
If you then differentiate you get
dy/dx = -2x - 6
for any point on this curve
2006-08-20 00:06:57 · answer #7 · answered by dope_move_busta 1 · 0⤊ 0⤋
At extreme values, the derivative is always zero.
At other values, calculate the slope of the curve at that point... this is the derivative at that point.
2006-08-19 16:01:49 · answer #8 · answered by DG 3 · 0⤊ 0⤋
Consult your logarithms!
2006-08-19 16:01:29 · answer #9 · answered by Anonymous · 0⤊ 0⤋