# At every point of the graph of the function y=f(x) we have y''=30x and at the point (0,3)The tangent to the graph of y = f(x) is parallel to the line 9x-4y+4 = 0. Find the rule for f(x).

### 1 Answer | Add Yours

Notice that the second order derivative is a polynomial of first order, hence, the function needs to be a third order polynomial such that:

`f(x) = ax^3 + bx^2 + cx + d`

Differentiating f(x) yields:

`f'(x) = 3ax^2 + 2bx + c`

Differentiating `f'(x)` yields:

`f''(x) = 6ax + 2b`

Notice that the problem provides the information that `f''(x) = 30x` , hence, equating `30x` and `6ax + 2b` yields:

`6ax + 2b = 30x`

Equating the coefficients of like powers yields:

`6a = 30 => a = 5`

`2b = 0 => b = 0`

The problem provides the information that the tangent line to the graph, at the point (0,3) is parallel to the line `9x-4y+4 = 0` such that:

`y - f(3) = f'(3)*x`

Since `f(3) = 0 => y = f'(3)*x`

You should remember that the slopes of two parallel lines are equal, hence, you need to convert the general form of equation of parallel line is slope intercept form such that:

`-4y = -9x - 4 => y = (9/4)x + 1`

Hence, the slope of the parallel line is `m = 9/4` .

You also should remember that `m = f'(3), ` hence `f'(3) = 9/4.`

You need to evaluate `f'(3)` substituting 3 for x in equation of f'(x) such that:

`f'(3) = 3a*9 + 2b*3 + c => 27a + 6b + c = 9/4`

Substituting 5 for a and 0 for b yields:

`27*5 + 0 + c = 9/4 => c = 9/4 - 135 => c = (9 - 540)/4`

`c = -531/4`

You may evaluate the coefficient d, substituting 5 for a, 0 for b and `-531/4` for c in equation `f(3) = 0` such that:

`5*27+0- 1593/4+ d = 0 => d = 1593/4 - 135`

`d = (1539 - 540)/4 => d = 999/4`

**Hence, evaluating the equation of the function f(x), under the given conditions, yields `f(x) = 5x^3 - -531/4x + 999/4` .**