Thread: Finding the area of a region

let f be the function given by f(x)=(4x^2)-x^3, and let l be the line y=18-3x, where l is tangent to the graph of f. Let D be the region bounded by the graph of f and the x-axis, and let E be the region bounded by the graph f, the line l, and the x-axis.

a. show that l is tangent to the graph of y=f(x) at the point x=3

Answer:
df/dx = 8x - 3x²
is the slope of the curve, which at x = 3 produces df/dx = 24 - 27 = -3
When x = 3, f(x) = 36 - 27 = 9.
The equation of a line passing through (3,9) with slope -3 is
y = -3x + b
9 = -3(3) + b
b = 18
→ y = 18 - 3x ← (a)


thats correct right?

You only need to show that f'(x) = y' evaluated at x = 3

since you are done.

Oh well that makes sense.

Originally Posted by jakncoke

You only need to show that f'(x) = y' evaluated at x = 3

since you are done.

It's more than that. You have to show the slopes are equal - so f'(3), the slope of the curve at x=3, equals -3, the slope of the line. But you also have to show that they pass through the same point there, so f(3), the y-value of the curve, equals 18-(3)(3), the y-value of the line.

The line y=17-3x matches slopes with the curve at x=3, but it is not tangent.

- Hollywood

Originally Posted by hollywood

It's more than that. You have to show the slopes are equal - so f'(3), the slope of the curve at x=3, equals -3, the slope of the line. But you also have to show that they pass through the same point there, so f(3), the y-value of the curve, equals 18-(3)(3), the y-value of the line.

The line y=17-3x matches slopes with the curve at x=3, but it is not tangent.

- Hollywood

yes you are absolutely right, for the slope or, derivative at a point is just a vector.