# Use "u substitution" to get answer f''(x) = 6x and the line y = 5-3x is tangent to the curve at the point where x = 1.

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### 2 Answers

Given `f''(x)=6x` and the line `y=5-3x` is tangent to the curve when x=1:

(1) If x=1 then `y=5-3(1)=2` so the function goes through the point (1,2) and `f'(1)=-3` as the first derivative gives the slope of the tangent line at a point.

(2) `f''(x)=6x ==> f'(x)=3x^2+C_1` . Since `f'(1)=-3` we have:

`3(1)^2+C_1=-3 ==> C_1=-6`

Thus `f'(x)=3x^2-6`

(3) The function contains the point (1,2) and

`f'(x)=3x^2-6 ==>f(x)=x^3-6x+C_2` `f(1)=2` so:

`(1)^3-6(1)+C_2=2`

`C_2=7`

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The function is `f(x)=x^3-6x+7`

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The graph of the function and the given tangent line:

f''(x) = 6x

f'(x)

`= int f''(x) dx`

`= int 6x dx`

Let ;

`U = 3x^2`

dU = 6xdx

f'(x)

`= int dU`

= U+C where C is a constant

`= 3x^2+C`

For any f(x) the gradient of the tangent line will be given by f'(x).

It is given that tangent to f(x) is y = 5-3x at x=1

So the gradient is -3

`f'(x) =3x^2+C `

since f'(x) is gradient

`3x^2+C = -3` at x=1

`3(1)^2+C = -3`

` C = -6`

so

`f'(x) = 3x^2-6`

f(x)

`= int f'(x) dx`

`= int (3x^2-6) dx`

Let;

`V = x^3-6x`

`dV = (3x^2-6) dx`

f(x)

`= int (3x^2-6) dx`

`= int dV`

`= V+C1`

`= x^3-6x+C1`

at x=1 f(x) and y=5-3x is tangent to each other.

`x^3-6x+C1 = 5-3x` at x =1

` `

`1^3-6*1+C1 = 5-3*1`

`C1 = 7`

**`f(x) = x^3-6x+7` **