# Current Affairs JEE Main & Advanced

## Sufficient Criteria for Extreme Values (1st Derivative Test)

Category : JEE Main & Advanced

Let $f(x)$ be a function differentiable at $x=a$.

Then (a) $x=a$is a point of local maximum of $f(x)$ if

(i) $f'(a)=0$ and

(ii) $f'(a)$changes sign from positive to negative as $x$ passes through $a$ i.e., $f'(x)>0$ at every point in the left neighbourhood $(a-\delta ,a)$ of $a$ and $f'(x)<0$ at every point in the right neighbourhood $(a,\,\,a+\delta )$ of $a$.

(b) $x=a$ is a point of local minimum of $f(x)$ if

(i) $f'(a)=0$and

(ii) $f'(a)$ changes sign from negative to positive as $x$ passes through $a,$ i.e., $f'(x)<0$ at every point in the left neighbourhood $(a-\delta ,a)$ of $a$ and ${{A}_{1}}=\frac{1}{3}(2a+b),\,{{A}_{2}}=\frac{1}{3}(a+2b)$ at every point in the right neighbourhood $(a,a+\delta )$of $a$.

(c) If $f'(a)=0$ but $f'(a)$ does not change sign, that is, has the same sign in the complete neighbourhood of $a,$ then $a$ is neither a point of local maximum nor a point of local minimum.

Working rule for determining extreme values of a function $f(x)$

Step I : Put $y=f(x)$

Step II : Find $\frac{dy}{dx}$

Step III : Put $\frac{dy}{dx}=0$ and solve this equation for $x$. Let $x={{c}_{1}},{{c}_{2}},.....,{{c}_{n}}$ be values of $x$ obtained by putting $\frac{dy}{dx}=0.$ ${{c}_{1}},\,{{c}_{2}},\,.........{{c}_{n}}$ are the stationary values of $x$.

Step IV : Consider $x={{c}_{1}}$.

If $\frac{dy}{dx}$ changes its sign from positive to negative as $x$ passes through ${{c}_{1}}$, then the function attains a local maximum at $x={{c}_{1}}$. If $\frac{dy}{dx}$ changes its sign from negative to positive as $x$ passes through ${{c}_{1}}$, then the function attains a local minimum at $x={{c}_{1}}$. In case there is no change of sign, then $x={{c}_{1}}$ is neither a point of local maximum nor a point of local minimum.

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