# Find the intervals of concavity and the points of inflexion of the function

\[f\left( x \right)={{\left( x-1 \right)}^{\dfrac{1}{3}}}\]

Last updated date: 24th Mar 2023

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Answer

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Hint: Double differentiate the given f(x) and find whether it is positive or negative. When f''(x) > 0, then the graph of f (x) is concave upwards and when f''(x )< 0, then the graph of f (x) is concave downwards. Also we will get the inflexion point at x where f'' (x) = 0 or does not exist.

Complete step-by-step answer:

Here we have to find the intervals of concavity and points of inflexion of function

\[f\left( x \right)={{\left( x-1 \right)}^{\dfrac{1}{3}}}\]

We know that if double derivative of function which is \[{{f}^{''}}\left( x \right)>0\text{ }\forall \text{ }x\in \left( a,b \right)\], then the graph of f (x) is concave upwards in (a, b).

Similarly, if \[{{f}^{''}}\left( x \right)<0\text{ }\forall \text{ }x\in \left( a,b \right)\], then the graph of f (x) is concave downwards in (a, b).

The point at which concavity of graph changes, i.e. \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=0\] or does not exist is the point of inflexion.

The sign of \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}\] changes about the point of inflexion.

Now, we can take

\[f\left( x \right)={{\left( x-1 \right)}^{\dfrac{1}{3}}}\]

We know that,

\[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\]

Now, we will differentiate f (x). We get,

\[{{f}^{'}}\left( x \right)=\dfrac{1}{3}{{\left( x-1 \right)}^{\dfrac{1}{3}-1}}\]

\[{{f}^{'}}\left( x \right)=\dfrac{1}{3}{{\left( x-1 \right)}^{\dfrac{-2}{3}}}\]

Now, again we will differentiate the above equation. We get,

\[{{f}^{''}}\left( x \right)=\dfrac{1}{3}\left( \dfrac{-2}{3} \right){{\left( x-1 \right)}^{\dfrac{-2}{3}-1}}\]

\[{{f}^{''}}\left( x \right)=\dfrac{-2}{9}{{\left( x-1 \right)}^{\dfrac{-5}{3}}}\]

Or, \[{{f}^{''}}\left( x \right)=\dfrac{-2}{9{{\left( x-1 \right)}^{\dfrac{5}{3}}}}\]

We know that the point of inflexion is a point where \[{{f}^{''}}\left( x \right)=0\] or does not exist. It changes its sign.

As we can see that \[{{f}^{''}}\left( x \right)\] cannot be zero but \[{{f}^{''}}\left( x \right)\] will not exist at x = 1 as (x – 1) which is in the denominator will become zero at x = 1.

Now, for x > 1, (x – 1) would be greater than zero. So,

\[{{\left( x-1 \right)}^{\dfrac{5}{3}}}>0\]

Therefore,

\[{{f}^{''}}\left( x \right)=\dfrac{-2}{9{{\left( x-1 \right)}^{\dfrac{5}{3}}}}<0\]

As for any value of x, say x = 2 in this case, we would get the value of \[{{f}^{''}}\left( x \right)\text{ as }\dfrac{-2}{9}\].

Now, for x < 1, (x – 1) would be less than zero. So,

\[{{\left( x-1 \right)}^{\dfrac{5}{3}}}<0\]

Therefore,

\[{{f}^{''}}\left( x \right)=\dfrac{-2}{9{{\left( x-1 \right)}^{\dfrac{5}{3}}}}>0\]

As for any value of x, say x = 0 in this case, we would get the value of \[{{f}^{''}}\left( x \right)\text{ as }\dfrac{2}{9}\].

Therefore, as f”(x) does not exist at x = 1, and f”(x) is positive for x < 1 and negative for x > 1, that is f (x) changes its sign about the point x = 1. Therefore, x = 1 is inflexion point.

As we have proved that f”(x) > 0 for x < 1

Therefore, f(x) is concave upwards for \[\left( -\infty ,1 \right)\]

As we have proved that f”(x) < 0 for x > 1

Therefore, f(x) is concave downwards for \[\left( 1,\infty \right)\]

Note: Students should always check if the function is changing its sign at an inflexion point and not only rely upon \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=0\] or does not exist for finding the inflexion point.

Complete step-by-step answer:

Here we have to find the intervals of concavity and points of inflexion of function

\[f\left( x \right)={{\left( x-1 \right)}^{\dfrac{1}{3}}}\]

We know that if double derivative of function which is \[{{f}^{''}}\left( x \right)>0\text{ }\forall \text{ }x\in \left( a,b \right)\], then the graph of f (x) is concave upwards in (a, b).

Similarly, if \[{{f}^{''}}\left( x \right)<0\text{ }\forall \text{ }x\in \left( a,b \right)\], then the graph of f (x) is concave downwards in (a, b).

The point at which concavity of graph changes, i.e. \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=0\] or does not exist is the point of inflexion.

The sign of \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}\] changes about the point of inflexion.

Now, we can take

\[f\left( x \right)={{\left( x-1 \right)}^{\dfrac{1}{3}}}\]

We know that,

\[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\]

Now, we will differentiate f (x). We get,

\[{{f}^{'}}\left( x \right)=\dfrac{1}{3}{{\left( x-1 \right)}^{\dfrac{1}{3}-1}}\]

\[{{f}^{'}}\left( x \right)=\dfrac{1}{3}{{\left( x-1 \right)}^{\dfrac{-2}{3}}}\]

Now, again we will differentiate the above equation. We get,

\[{{f}^{''}}\left( x \right)=\dfrac{1}{3}\left( \dfrac{-2}{3} \right){{\left( x-1 \right)}^{\dfrac{-2}{3}-1}}\]

\[{{f}^{''}}\left( x \right)=\dfrac{-2}{9}{{\left( x-1 \right)}^{\dfrac{-5}{3}}}\]

Or, \[{{f}^{''}}\left( x \right)=\dfrac{-2}{9{{\left( x-1 \right)}^{\dfrac{5}{3}}}}\]

We know that the point of inflexion is a point where \[{{f}^{''}}\left( x \right)=0\] or does not exist. It changes its sign.

As we can see that \[{{f}^{''}}\left( x \right)\] cannot be zero but \[{{f}^{''}}\left( x \right)\] will not exist at x = 1 as (x – 1) which is in the denominator will become zero at x = 1.

Now, for x > 1, (x – 1) would be greater than zero. So,

\[{{\left( x-1 \right)}^{\dfrac{5}{3}}}>0\]

Therefore,

\[{{f}^{''}}\left( x \right)=\dfrac{-2}{9{{\left( x-1 \right)}^{\dfrac{5}{3}}}}<0\]

As for any value of x, say x = 2 in this case, we would get the value of \[{{f}^{''}}\left( x \right)\text{ as }\dfrac{-2}{9}\].

Now, for x < 1, (x – 1) would be less than zero. So,

\[{{\left( x-1 \right)}^{\dfrac{5}{3}}}<0\]

Therefore,

\[{{f}^{''}}\left( x \right)=\dfrac{-2}{9{{\left( x-1 \right)}^{\dfrac{5}{3}}}}>0\]

As for any value of x, say x = 0 in this case, we would get the value of \[{{f}^{''}}\left( x \right)\text{ as }\dfrac{2}{9}\].

Therefore, as f”(x) does not exist at x = 1, and f”(x) is positive for x < 1 and negative for x > 1, that is f (x) changes its sign about the point x = 1. Therefore, x = 1 is inflexion point.

As we have proved that f”(x) > 0 for x < 1

Therefore, f(x) is concave upwards for \[\left( -\infty ,1 \right)\]

As we have proved that f”(x) < 0 for x > 1

Therefore, f(x) is concave downwards for \[\left( 1,\infty \right)\]

Note: Students should always check if the function is changing its sign at an inflexion point and not only rely upon \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=0\] or does not exist for finding the inflexion point.

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