Question

Let $f\left(x\right)$ be a polynomial function such that $f\left(x\right)+f^{\prime }\left(x\right)+f^{\prime \prime }\left(x\right)=x^5+64$. Then, the value of $\underset{x\to 1}{lim}\frac{f\left(x\right)}{x-1}$ is equal to:

Answer 1

$\underset{x\to 1}{lim}\frac{f\left(x\right)}{x-1}$
$f\left(x\right)+f^{\prime }\left(x\right)+f^{\prime \prime }\left(x\right)=x^5+64$
Let $f\left(x\right)=x^5+a{x}^4+b{x}^3+c{x}^2+dx+e$
$f^{\prime }\left(x\right)=5x^4+4a{x}^3+3b{x}^2+2cx+d$
$f^{\prime \prime }\left(x\right)=20x^3+12a{x}^2+6bx+2c$
$x^5+(a+5)x^4+(b+4a+20)x^3+(c+3b+12a)x^2+(d+2c+6b)x+e+d+2c=x^5+64$
$\Rightarrow a+5=0$
$b+4a+20=0$
$c+3b+12a=0$
$d+2c+6b=0$
$e+d+2c=64$
$\therefore a=-5,b=0,c=60,d=-120,e=64$
$\therefore f\left(x\right)=x^5-5x^4+60x^2-120x+64$
Now, $\underset{x\to 1}{lim}\frac{x^5-5x^4+60x^2-120x+64}{x-1}$ is $\left(\frac{0}{0}\text{from}\right)$
By L' Hopital rule
$\underset{x\to 1}{lim}\frac{5x^4-20x^3+120x-120}{1}$
$=-15$