Need help on PRELIM calculus Q (1 Viewer)
- Thread starter Marc26
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the first thing would be to differentiate the curve, so we would get dy/dx = -x^2
when we substitute values for x into dy/dx, we are basically finding the gradient of the tangent to the curve curve at that particular x-value, for example, in this case, if we were to substitute x=3 into dy/dx, we would find the gradient of the tangent to the curve y= 3 - 1/3*x^3 at the point x=3
knowing this, if we let dy/dx= -1, and solve for x, we should be able to find the values of x for which the tangent's gradient is equal to -1
so when dy/dx = -1,
-x^2 = -1
x^2 = 1
x = 1 or -1
now we know that the tangents to the curve y = 3 - 1/3*x^3 at x=1 and x=-1 have a gradient of -1
substituting the x-values into the curve, we get the points (1, 8/3) and (-1, 10/3)
so at these 2 points, their tangents to the curve have a gradient of -1
when we substitute values for x into dy/dx, we are basically finding the gradient of the tangent to the curve curve at that particular x-value, for example, in this case, if we were to substitute x=3 into dy/dx, we would find the gradient of the tangent to the curve y= 3 - 1/3*x^3 at the point x=3
knowing this, if we let dy/dx= -1, and solve for x, we should be able to find the values of x for which the tangent's gradient is equal to -1
so when dy/dx = -1,
-x^2 = -1
x^2 = 1
x = 1 or -1
now we know that the tangents to the curve y = 3 - 1/3*x^3 at x=1 and x=-1 have a gradient of -1
substituting the x-values into the curve, we get the points (1, 8/3) and (-1, 10/3)
so at these 2 points, their tangents to the curve have a gradient of -1
