Question

In problems 45 and 46 use problem 55 in Exercises 1.1 and (2) and (3) of this section.

Find a function whose second derivative is $\mathbf{y}\text{'}\text{'}=\mathbf{12}\mathbf{x}-\mathbf{2}$at each point (x, y) on its graph and $\mathbf{y}=-\mathbf{x}+\mathbf{5}$is tangent to the graph at the point corresponding to $\mathbf{x}=\mathbf{1}$.

Short answer

$y=2x^3-x^2-5x+8$.

Step-by-step solution

Slope of the tangent

The slope of the tangent line to a curve at a given point is equal to the slope of the function at that point, and the derivative of a function tells us its slope at any point.

Integration to calculate y'

First, let’s integrate to calculate$y\text{'}$, to get the slope of the tangent to the equation at any point :

$\begin{array}{c}\int y"=\int (12x-2)dx\\ =\int 12xdx-\int 2dx\\ =12\int xdx-2\int dx\end{array}$

Integrating we get,

$$\begin{gathered} y\text{'}=12·\frac{x^2}{2}-2x+C_1 \\ y\text{'}=6x^2-2x+C_1 \end{gathered}$$

Find the value of constant C

Besides, we also have information that $y=-x+5$is tangent to the graph at the point corresponding to $x=1$, which implies that the slope is $-1$and the initial condition is: $y\text{'}\left(1\right)=-1$.

Therefore,

Substitute $y\text{'}\left(1\right)=-1$in the equation $y\text{'}=6x^2-2x+C_1$

$$\begin{gathered} 6(1{)}^2-2\left(1\right)+C_1=-1 \\ 6-2+C_1=-1 \\ {C}_1=-5 \end{gathered}$$

So,$y\text{'}=6x^2-2x-5$

Step 4:Integrating to find y and another constant

It still remains to find y:

So, integrate y' to get y:

$\begin{array}{l}\int y\text{'}dx=\int (6x^2-2x-5)dx\\ \int y\text{'}dx=\int 6x^2-\int 2x-\int 5dx\end{array}$

Integrating we get,

$$\begin{gathered} y=6\left(\frac{x^3}{3}\right)-2\left(\frac{x^2}{2}\right)-5x+C_2 \\ y=2x^3-x^2-5x+C_2 \end{gathered}$$

Now use the fact that graph passes through the point $(1,4)$and calculate $C_2$.

Remember that the equation of the tangent is $y=-x+5$and when $x=1,y=4$.

Substitute the values, we get

$$\begin{gathered} y\left(1\right)=2(1{)}^3-(1{)}^2-5\left(1\right)+C_2 \\ 4=2-1-5+C_2 \end{gathered}$$

Simplify,

$C_2=8$

Therefore, the solution is.$y=2x^3-x^2-5x+8$.