Question

Let $f:\left[ {-1,1} \right] \to R$ be defined as $f\left( x \right) = a{x^2} + bx + c$ for all $x \in [-1,1]$, where $a,b,c \in R$ such that $f\left( -1 \right)=2$, $f’\left( -1 \right)=1$ and for $x \in [-1,1]$ the maximum value of $f’’\left(x\right)$ is $\dfrac{1}{2}$ . If $f\left(x\right)\leq \alpha$, $x \in [-1,1]$, then the least value of $\alpha$ is equal to.

Answer 1

Hint: Differentiate the given function $f\left( x \right) = a{x^2} + bx + c$. So, we can equate it with the maximum value of that function given and determine the value of constant. Thereafter substituting the obtained value and equating it with the first differentiation to evaluate the value of other constants in the function. Finally putting obtained values in the function and equating it with the given value to know the least value.

Formula Used:
$\dfrac{d}{dx}x^{n}=nx^{n-1}$
$\dfrac{d}{dx}\left[m f(x)\right]=m\dfrac{d}{dx}\left[ f(x)\right]$

Complete step by step solution:
The given quadratic equation $f\left( x \right) = a{x^2} + bx + c$ ………. (1)
Differentiate it with respect to x, we get
  $f\left( x \right)=2ax+b$ …….. (2)
Again, differentiate it with respect to x, we get
  $f\left( x \right)=2a$ ……… (3)
Evaluate the obtained equation with given values.
$f\left( 1 \right)=\dfrac{1}{2}$ ……… (4)
Equating equation (4) with (3), we get
$ \Rightarrow a = \dfrac{1}{4}$
Evaluate the obtained equation with given values.
$f\left( 1 \right)=1$ ……… (5)
Equating equation (5) with (2), we get
$ \Rightarrow b-2a = 1$
Evaluate it further, we get
$ \Rightarrow b = \dfrac{3}{2}$
Evaluate the obtained equation with given values.
$f\left( {-1} \right) = a-b + c = 2$
Substituting the obtained values of a and b in above equation, we get
$ \Rightarrow c = \dfrac{{13}}{4}$
Now substitute the value of a, b and c in the quadratic equation given.
$f\left( x \right) = \left( {\dfrac{1}{4}} \right)\left( {{x^2} + 6x + 13} \right),x \in [-1,1]$
Derivative of the above function will be:
$f\left( x \right)=\left( \dfrac{1}{4} \right)\left( 2x+6 \right)=0$
Evaluate the value of x
$ \Rightarrow x = -3 \notin [-1,1]$
Then the end values of function will be:
$f\left( 1 \right) = 5,f\left( {-1} \right) = 2$
So, the function will take the value
$f\left( x \right) \le 5$
Thus, ${\alpha _{minimum}} = 5$

Note: By using the initial conditions $f\left( -1 \right)=2$, $f’\left( -1 \right)=1$ and $f’’\left(x\right)= \dfrac{1}{2}$,we will calculate the value of a,b,c of the given function $f\left( x \right) = a{x^2} + bx + c$. Given that $x\in \left[ -1,1 \right]$. We will put $x = -1$ and $x = 1$ in the given function. From this we will find the maximum value of $f\left( x \right)$ and $\alpha$.