Question

Find a function y = f(x) 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 .

Answer 1

integrate 12x - 2 to gety'
y'= 6x^2 - 2x + c
integrate again to get y

y = 2x^3 - x^2 + cx + d where c and d are constants

Answer 2

the slope of the tangent = f'

= 6x^2 - 2x + c
at x = 1

this is 6 - 2 + c
the tangent equation is y = -x + 5

so slope = -1 = 6-2+c
giving c = -1 -6+ 2 = -5

giving f' = 6x^2 -2x -5
and fx) = 2x^3 -x^2 - 5x + d

now we need to find value of d

Answer 3

tangent equation is
y = -x + 5
when x = 1
y = -1 + 5 = 4

this point is also on graph of f(x) so

4 = 2(1)^3 -1^2 - 5(1) + d
solve this for d and plug this into the expression we got in last post