# Investigate\[\left\{ \begin{matrix}

{{x}^{3}}+{{x}^{2}}+10x, & x<0 \\

-3\sin x, & x\ge 0 \\

\end{matrix} \right.\], at x=0 for local maxima/minima.

Last updated date: 19th Mar 2023

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Answer

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304.8k+ views

Hint: First confirm that function is continuous at x = 0 or not. Then find the left hand and right hand derivatives of the function and apply the fact that if the left hand derivative is greater than zero and right hand derivative is less than zero then it is a local maxima. If the left hand derivative is less than zero and the right hand derivative is greater than then it is a local minima. Accordingly, get the desired result.

Complete step-by-step answer:

At first will prove whether f(x) is continuous at x = 0 by finding the left and right hand limit.

If \[x\to {{0}^{-}}\] then we have to consider f(x) as \[{{x}^{3}}+{{x}^{2}}+10x\].

So, \[f\left( {{0}^{-}} \right)={{0}^{3}}+{{0}^{2}}+10\left( 0 \right)=0\].

If \[x\to {{0}^{+}}\], then we have to consider f(x) as -3sin x.

So, \[f\left( {{0}^{+}} \right)=-3\sin 0=0\].

If \[x\to {{0}^{+}}\] then the value of \[f\left( {{0}^{+}} \right)\] is considered as right hand limit and if \[x\to {{0}^{-}}\] then the value of

\[f\left( {{0}^{-}} \right)\]is considered as a left hand limit.

So, we found out that the left hand and right hand limit both are equal to 0.

Therefore, f(x) is continuous.

Now for checking the point as local minima or maxima we have to find its left hand and right hand derivative, if left hand derivative is negative and right hand derivative is positive then, it is local minima and if left hand derivative is positive and right hand derivative is negative then it is maxima.

Let’s first find the left hand derivative when \[x\to {{0}^{-}}\] then,

\[f\left( x \right)={{x}^{3}}+{{x}^{2}}+10\] then,

\[f'\left( x \right)=3{{x}^{2}}+2x+10\]

Here we have used the identity \[\dfrac{d}{dx}{{x}^{n}}=n{{x}^{-1}}\], where n is any constant.

So, \[f'\left( {{0}^{-}} \right)=3{{\left( 0 \right)}^{2}}+2\left( 0 \right)+10=10\]

So, the value of the left hand derivative is 10.

Now let’s find the right hand derivative when \[x\to {{0}^{+}}\] then,

\[f\left( x \right)=-3\sin x\]

Then, \[f\left( x \right)=-3\cos x\]

Here we have used the identity \[\dfrac{d}{dx}\sin x=\cos x\].

Now, \[f\left( {{0}^{+}} \right)=-3\cos 0=-3\]

So, the value of the right hand derivative is -3.

Now we found out that left hand derivative is positive and right hand derivative is negative so the point is considered to be local maxima.

Therefore at point ‘0’ the function is maxima.

Note: While finding left hand derivative we have to find value of function when \[x\to {{0}^{-}}\] and when finding value the right hand derivative, we have to find the value of function when \[x\to {{0}^{+}}\]. Also be careful about taking the function while finding values because this is the place where students make mistakes.

Complete step-by-step answer:

At first will prove whether f(x) is continuous at x = 0 by finding the left and right hand limit.

If \[x\to {{0}^{-}}\] then we have to consider f(x) as \[{{x}^{3}}+{{x}^{2}}+10x\].

So, \[f\left( {{0}^{-}} \right)={{0}^{3}}+{{0}^{2}}+10\left( 0 \right)=0\].

If \[x\to {{0}^{+}}\], then we have to consider f(x) as -3sin x.

So, \[f\left( {{0}^{+}} \right)=-3\sin 0=0\].

If \[x\to {{0}^{+}}\] then the value of \[f\left( {{0}^{+}} \right)\] is considered as right hand limit and if \[x\to {{0}^{-}}\] then the value of

\[f\left( {{0}^{-}} \right)\]is considered as a left hand limit.

So, we found out that the left hand and right hand limit both are equal to 0.

Therefore, f(x) is continuous.

Now for checking the point as local minima or maxima we have to find its left hand and right hand derivative, if left hand derivative is negative and right hand derivative is positive then, it is local minima and if left hand derivative is positive and right hand derivative is negative then it is maxima.

Let’s first find the left hand derivative when \[x\to {{0}^{-}}\] then,

\[f\left( x \right)={{x}^{3}}+{{x}^{2}}+10\] then,

\[f'\left( x \right)=3{{x}^{2}}+2x+10\]

Here we have used the identity \[\dfrac{d}{dx}{{x}^{n}}=n{{x}^{-1}}\], where n is any constant.

So, \[f'\left( {{0}^{-}} \right)=3{{\left( 0 \right)}^{2}}+2\left( 0 \right)+10=10\]

So, the value of the left hand derivative is 10.

Now let’s find the right hand derivative when \[x\to {{0}^{+}}\] then,

\[f\left( x \right)=-3\sin x\]

Then, \[f\left( x \right)=-3\cos x\]

Here we have used the identity \[\dfrac{d}{dx}\sin x=\cos x\].

Now, \[f\left( {{0}^{+}} \right)=-3\cos 0=-3\]

So, the value of the right hand derivative is -3.

Now we found out that left hand derivative is positive and right hand derivative is negative so the point is considered to be local maxima.

Therefore at point ‘0’ the function is maxima.

Note: While finding left hand derivative we have to find value of function when \[x\to {{0}^{-}}\] and when finding value the right hand derivative, we have to find the value of function when \[x\to {{0}^{+}}\]. Also be careful about taking the function while finding values because this is the place where students make mistakes.

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