Question

Let $f\left(x\right)$ be a polynomial of degree $3$such that $f(-1)=10,f\left(1\right)=-6$ has a critical point at $x=-1$and $f\text{'}\left(x\right)$ has a critical point at $x=1$.Then the local minima at $x=?$.

Answer 1

Finding local minima at $x$.

Let the polynomial of degree $3$ be $a{x}^3+b{x}^2+cx+d$.

Given, $f(-1)=10,f\left(1\right)=-6$

Put the values $x=-1\text{and}1$in polynomial and compare.

$$\begin{gathered} \Rightarrow f(-1)=a{(-1)}^3+b{(-1)}^2+c(-1)+d \\ \Rightarrow 10=-a+b-c+d\dots \left(1\right) \\ \Rightarrow f\left(1\right)=a{\left(1\right)}^3+b{\left(1\right)}^2+c\left(1\right)+d \\ \Rightarrow -6=a+b+c+d\dots \left(2\right) \end{gathered}$$

Add equation $\left(1\right)\text{and}\left(2\right)$, we get,

$$\begin{gathered} \Rightarrow 2b+2d=10-6 \\ \Rightarrow 2(b+d)=4 \\ \Rightarrow b+d=2\dots \left(3\right) \end{gathered}$$

Subtract equation $\left(2\right)\text{from}\left(1\right)$, we get,

$$\begin{gathered} \Rightarrow -2a-2c=10-(-6) \\ \Rightarrow -2(a+c)=16 \\ \Rightarrow a+c=-8\dots \left(4\right) \end{gathered}$$

Now differentiating the polynomial with respect to $x$.

$\Rightarrow f\text{'}\left(x\right)=3a{x}^2+2bx+c$

Given, $f\text{'}(-1)=0$ as it is a critical point.

Substitute $x=-1$, we get

$$\begin{gathered} \Rightarrow f\text{'}(-1)=3a{\left(-1\right)}^2+2b\left(-1\right)+c \\ \Rightarrow 0=3a-2b+c\dots \left(5\right) \end{gathered}$$.

Now differentiating $f\text{'}\left(x\right)$ with respect to $x$.

$\Rightarrow f\text{'}\text{'}\left(x\right)=6ax+2b$

Given, $f\text{'}\text{'}\left(1\right)=0$

Substituting $x=1$.

$$\begin{gathered} \Rightarrow f\text{'}\text{'}\left(1\right)=6a\left(1\right)+2b \\ \Rightarrow 0=6a+2b \\ \Rightarrow b=-3a\dots \left(6\right) \end{gathered}$$

Put the value of $b\text{in}\left(5\right)$, we get,

$c=3b\dots \left(7\right)$

Putting value of $\left(6\right)\text{and}\left(7\right)in\left(4\right)$, we get,$a=1,b=-3,c=-9$

Also from equation $\left(3\right)$, we get $d=5$

Hence, the required polynomial is $x^3-3x^2-9x+5$

Differentiating polynomial with respect to $x$, we get

$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)=3x^2-6x-9 \\ \Rightarrow 0=3x^2-6x-9(f\text{'}(x)=0,\text{for checking its critical point}) \\ \Rightarrow 0=(x-3)(x+1) \end{gathered}$$

So, $x=-1,3$

Also, $f\text{'}\text{'}\left(x\right)=6x-6$

For $x=3,f\text{'}\text{'}\left(x\right)>0$

Hence, $f\left(x\right)$ has a local minima at $x=3$