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

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(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 $[1, 2]$?

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