Question

Compute the value of $\theta$ in the first mean value theorem $f(x+h)=f\left(x\right)+h{f}^{\prime }(x+\theta h)$ if $f\left(x\right)=a{x}^2+bx+c.$

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{5}$

Answer 1

The correct option is A $\frac{1}{2}$

We have $f\left(x\right)=a{x}^2+bx+c.$
$\Rightarrow f(x+h)=a{(x+h)}^2+b(x+h)+c$
Also, $f^{\prime }\left(x\right)=2ax+b$
$\Rightarrow f^{\prime }(x+\theta h)=2a(x+\theta h)+c$
Putting these values in Lagrange's mean value theorem we get
$\therefore f(x+h)=f\left(x\right)+h{f}^{\prime }(x+\theta h)a{(x+h)}^2+b(x+h)+c=a{x}^2+bx+c+h[2a(x+\theta h)+b]$
when $x\to 0.$ we have $a{h}^2+bh+c+h[2a\theta h+b]$
$a{h}^2=2a\theta h^2$ or $\theta =\frac{1}{2}$
which is the required value of $\theta .$