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Stella [2.4K]
3 years ago
9

find a function whose second derivative is y''=12x-2 at each point (x,y) on its graph and y=-x+5 is tangent to the graph at the

point corresponding to x=1
Mathematics
1 answer:
pogonyaev3 years ago
3 0
Start by integrating y'' twice to get function y(x).

y' = \int (12x - 2) dx = 6x^2 -2x + c \\  \\ y = \int (6x^2 -2x+c )dx= 2x^3-x^2 +cx+d

Next find value of coefficients 'c' and 'd' using the given tangent line
The slope of tangent line is equal to y' when x=1.

6x^2 -2x+c = -1, x = 1 \\  \\ 6-2+c = -1 \\  \\ c = -5

The tangent line intersects the curve y(x) when x = 1.
If x = 1, then y = 4  (From y =-x+5)

y = 2x^3 -x^2-5x +d , x = 1, y = 4 \\  \\ 2-1-5+d = 4 \\  \\ d = 8

Final Answer:
y = 2x^3 -x^2 -5x+8
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