For what values of a is the function f given | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For a function $f(x)$ to be increasing on an interval,its derivative $f'(x)$ must be greater than or equal to $0$ for all $x$ in that interval.Giv

Vedclass · Accepted Answer

$a \ge -2$

Vedclass · Answer

(A) For a function $f(x)$ to be increasing on an interval,its derivative $f'(x)$ must be greater than or equal to $0$ for all $x$ in that interval.Given $f(x) = x^2 + ax + 1$,the derivative is $f'(x) = 2x + a$.For $f(x)$ to be increasing on $[1, 2]$,we require $f'(x) \ge 0$ for all $x \in [1, 2]$.This implies $2x + a \ge 0$ for all $x \in [1, 2]$.Since $2x + a$ is an increasing linear function,its minimum value on the interval $[1, 2]$ occurs at the smallest value of $x$,which is $x = 1$.Therefore,we need $f'(1) \ge 0$.Substituting $x = 1$ into the derivative: $2(1) + a \ge 0$.$2 + a \ge 0$,which gives $a \ge -2$.Thus,the function is increasing on $[1, 2]$ for $a \ge -2$.

For what values of $a$ is the function $f$ given by $f(x) = x^2 + ax + 1$ increasing on the interval $[1, 2]$?

Similar Questions

Prove that the function given by $f(x) = x^{3} - 3x^{2} + 3x - 100$ is increasing in $R$.

For every value of $x$,the function $f(x) = \frac{1}{5^x}$ is:

Show that the function given by $f(x) = \sin x$ is strictly increasing in the interval $\left(0, \frac{\pi}{2}\right)$.

The function $f(x) = 1 - x^3 - x^5$ is a decreasing function for:

Let $f(x) = \frac{x}{\sqrt{a^2+x^2}} - \frac{d-x}{\sqrt{b^2+(d-x)^2}}$,$x \in R$,where $a, b, d$ are non-zero real constants. Then

Vedclass Test Series

Exam Paper Generator

Online Exam Module