Question

f (x) is a polynomial of the third degree which has a local maximum at x=−1. If f (1) =−1, f (2) = 18 and f ′(x) has a local minimum at x=0 then
A) f (0) = 5
B) f(x) has local minimum at x =1
C) f(x) is increasing in \[\left[ 1,2\sqrt{5} \right]\]
D) the distance between (-1, 2) and (a, f(a)), where a is point of local minimum is $2\sqrt{5}$

Answer 1

Hint: For solving this problem, first we assume a cubic polynomial and try to obtain the actual polynomial from the given conditions. By using the condition of maxima and minima we easily obtain the value of all the variables. After obtaining the final expression, we can check for all the correct options.

Complete step-by-step answer:
Let the cubic polynomial be $f\left( x \right)=a{{x}^{3}}+b{{x}^{2}}+cx+d$.
Whenever it is given that a function has a local maximum or minimum, at that particular value of x the derivative of that function is zero. We are given the local maximum at x = -1. Also, the derivative of f(x) has a local minimum at x = 0. We form two equations using this information as:
\[\begin{align}
  & {f}'{{\left( x \right)}_{x=-1}}=0 \\
 & \Rightarrow 3a{{x}^{2}}+2bx+c=0 \\
 & \Rightarrow 3a{{\left( -1 \right)}^{2}}+2b\left( -1 \right)+c=0 \\
 & \Rightarrow 3a-2b+c=0\ldots \left( 1 \right) \\
 & \text{Also, }{f}''{{\left( x \right)}_{x=0}}=0 \\
 & \Rightarrow 6ax+2b=0 \\
 & \Rightarrow 6a\left( 0 \right)+2b=0 \\
 & \Rightarrow b=0 \\
\end{align}\]
Also, the value of f (1) = -1 and f (2) = 18. So, another set of equations are:
$\begin{align}
  & f\left( 1 \right)=-1 \\
 & \Rightarrow a{{\left( 1 \right)}^{3}}+b{{\left( 1 \right)}^{2}}+c\left( 1 \right)+d=-1 \\
 & \Rightarrow a+b+c+d=-1 \\
 & \text{As, b=0} \\
 & \Rightarrow a+c+d=-1\ldots \left( 2 \right) \\
\end{align}$
As, from equation (1), 3a -2(0) + c = 0, so c = -3a. Now, substituting in equation (2), we get
$\begin{align}
  & \Rightarrow a-3a+d=-1 \\
 & \Rightarrow d-2a=-1\ldots \left( 3 \right) \\
\end{align}$
Again, f (2) = 18. Proceeding again, we get
$\begin{align}
  & f\left( 2 \right)=18 \\
 & \Rightarrow a{{\left( 2 \right)}^{3}}+b{{\left( 2 \right)}^{2}}+c\left( 2 \right)+d=18 \\
 & \Rightarrow 8a+4b+2c+d=18 \\
 & \text{As }b=0,\text{ }c=-3a\text{ and }d=2a-1 \\
 & \Rightarrow 8a+2\left( -3a \right)+2a-1=18 \\
 & \Rightarrow 8a-6a+2a-1=18 \\
 & \Rightarrow 4a-1=18 \\
 & \Rightarrow 4a=19 \\
 & \Rightarrow a=\dfrac{19}{4} \\
 & \Rightarrow d=2\left( \dfrac{19}{4} \right)-1 \\
 & \Rightarrow d=\dfrac{38-4}{4}=\dfrac{34}{4} \\
 & \Rightarrow c=\dfrac{-3\times 19}{4}=\dfrac{-57}{4} \\
\end{align}$
Now, the obtained function is $f\left( x \right)=\dfrac{1}{4}\left( 19{{x}^{3}}-57x+34 \right)$.
Now, obtaining f (0), we get
$\begin{align}
  & f\left( 0 \right)=\dfrac{1}{4}\left( 19\times 0-57\times 0+34 \right) \\
 & \therefore f\left( 0 \right)=\dfrac{34}{4}\ne 5 \\
\end{align}$
Therefore, option (a) is incorrect.
Now, differentiating with respect to x, we get
$\begin{align}
  & {f}'\left( x \right)=\dfrac{1}{4}\left( 57{{x}^{2}}-57 \right) \\
 & {f}'\left( x \right)=0 \\
 & \Rightarrow 57{{x}^{2}}-57=0 \\
 & \Rightarrow {{x}^{2}}-1=0 \\
 & \Rightarrow \left( x-1 \right)\left( x+1 \right)=0 \\
 & \Rightarrow x=1,-1 \\
 & {f}''\left( x \right)=\dfrac{1}{4}\left( 114x \right) \\
 & {f}''\left( x \right)>0\text{ at }x=1 \\
\end{align}$
Therefore, at x = 1, we have a local minimum. Therefore, option (b) is correct.
Now, for checking the function as increasing f’(x) > 0.
$\begin{align}
  & \text{As, }\left( x+1 \right)\left( x-1 \right)>0 \\
 & x\in \left( -\infty ,-1 \right]\cup \left[ 1,\infty \right) \\
\end{align}$
Therefore, option (c) is correct.
Now, for at x = 1, we have local minimum and the value of function at x = 1 is:
$\begin{align}
  & f\left( 1 \right)=\dfrac{1}{4}\left[ 19{{\left( 1 \right)}^{3}}-57\left( 1 \right)+34 \right] \\
 & f\left( 1 \right)=\dfrac{1}{4}\left( 19-57+34 \right) \\
 & f\left( 1 \right)=\dfrac{-4}{4}=-1 \\
\end{align}$
The distance between (-1, 2) and (1, -1) can be given by formula:
$\begin{align}
  & d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}} \\
 & d=\sqrt{{{\left( 1-\left( -1 \right) \right)}^{2}}+{{\left( -1-2 \right)}^{2}}} \\
 & d=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( -3 \right)}^{2}}} \\
 & d=\sqrt{4+9} \\
 & d=\sqrt{13} \\
\end{align}$
Therefore, $\sqrt{13}\ne 2\sqrt{5}$ option (d) is incorrect.
Hence, option (b) and (c) is correct.

Note: The key step in solving this problem is the formation of cubic polynomials by using the given information. After obtaining the function, the double derivative should be calculated carefully because the sign of double derivative gives the indication of existence of minima and maxima at critical points where the first derivative is zero. The correctness of obtained function can be easily checked by using the problem statement conditions.