Chapter 3.1, Problem 61E

To determine

To find: The second degree polynomial.

Answer to Problem 61E

The second degree polynomial is $\underset{\_}{P\left(x\right)=x^2-x+3}$.

Explanation of Solution

Given:

The second degree polynomial satisfies $P\left(2\right)=5$ ,$P^{\prime }\left(2\right)=3$ and $P^{″}\left(2\right)=2$.

Derivative rules:

(1) Constant multiple rule: $\frac{d}{dx}\left(cf\right)=c\frac{d}{dx}\left(f\right)$

(2) Power rule: $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

(3) Sum rule: $\frac{d}{dx}\left(f+g\right)=\frac{d}{dx}\left(f\right)+\frac{d}{dx}\left(g\right)$

Calculation:

The general form of the second degree polynomial is $P\left(x\right)=a{x}^2+bx+c$ (1)

Obtain the first derivative of $P\left(x\right)=a{x}^2+bx+c$.

$\begin{array}{c}{P}^{\prime }\left(x\right)=\frac{d}{dx}\left(P\left(x\right)\right)\\ =\frac{d}{dx}\left(a{x}^2+bx+c\right)\end{array}$

Apply the sum rule (3) and the constant multiple (1),

$\begin{array}{c}{P}^{\prime }\left(x\right)=\frac{d}{dx}\left(a{x}^2\right)+\frac{d}{dx}\left(bx\right)+\frac{d}{dx}\left(c\right)\\ =a\frac{d}{dx}\left(x^2\right)+b\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(c\right)\end{array}$

Since the derivative of constant function is zero, $\frac{d}{dx}\left(c\right)=0$.

$\begin{array}{c}{P}^{\prime }\left(x\right)=a\frac{d}{dx}\left(x^2\right)+b\frac{d}{dx}\left(x\right)+0\\ =a\frac{d}{dx}\left(x^2\right)+b\frac{d}{dx}\left(x\right)\end{array}$

Apply the power rule (2) and simplify the terms,

$\begin{array}{c}{P}^{\prime }\left(x\right)=a\left(2x^{2-1}\right)+b\left(1x^{1-1}\right)\\ =2a\left(x\right)+b\left(1\right)\\ =2ax+b\end{array}$

Therefore, the first derivative of $P\left(x\right)$ is $\underset{\_}{P^{\prime }\left(x\right)=2ax+b}$.

Obtain the second derivative of $P\left(x\right)=a{x}^2+bx+c$.

$\begin{array}{c}{P^{\prime }}^{\prime }\left(x\right)=\frac{d}{dx}\left(P^{\prime }\left(x\right)\right)\\ =\frac{d}{dx}\left(2ax+b\right)\end{array}$

Apply the sum rule (3) and the constant multiple (1),

$\begin{array}{c}{P^{\prime }}^{\prime }\left(x\right)=\frac{d}{dx}\left(2ax\right)+\frac{d}{dx}\left(b\right)\\ =2a\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(b\right)\end{array}$

Since the derivative of constant function is zero, $\frac{d}{dx}\left(b\right)=0$.

${P^{\prime }}^{\prime }\left(x\right)=2a\frac{d}{dx}\left(x\right)$

Apply the power rule (2) and simplify the terms,

$\begin{array}{c}{P^{\prime }}^{\prime }\left(x\right)=2a\left(1x^{1-1}\right)\\ =2a\end{array}$

Therefore, the second derivative of $P\left(x\right)$ is $\underset{\_}{P^{″}\left(x\right)=2a}$.

Obtain the second degree polynomial P.

Substitute 2 for x in $P^{″}\left(x\right)=2a$,

$P^{″}\left(2\right)=2a$

Since $P^{″}\left(2\right)=2$, $2a=2$.

Thus, the value of $a=1$.

Substitute 2 for x in $P^{\prime }\left(x\right)=2ax+b$,

$\begin{array}{c}{P}^{\prime }\left(2\right)=2a\left(2\right)+b\\ =4a+b\\ =4\left(1\right)+b\left[Qa=1\right]\\ =4+b\end{array}$

Since $P^{\prime }\left(2\right)=3$, $4+b=3$.

$\begin{array}{c}b=3-4\\ =-1\end{array}$

Thus, the value of $b=-1$ .

Substitute 2 for x in $P\left(x\right)=a{x}^2+bx+c$,

$\begin{array}{c}P\left(2\right)=a{\left(2\right)}^2+b\left(2\right)+c\\ =4a+2b+c\\ =4\left(1\right)+2\left(-1\right)+c\left[Qa=1,b=-1\right]\\ =2+c\end{array}$

Since $P\left(2\right)=5$, $2+c=5$.

$\begin{array}{c}c=5-2\\ =3\end{array}$

Thus, the value of $c=2$.

Substitute the values 1 for a,-1 for b and 3 for c in $P\left(x\right)=a{x}^2+bx+c$,

$\begin{array}{c}P\left(x\right)=\left(1\right)x^2+\left(-1\right)x+3\\ =x^2-x+3\end{array}$

Therefore, the second degree polynomial is $\underset{\_}{P\left(x\right)=x^2-x+3}$.