
How many points of inflection does the function \[f\left( x \right)={{x}^{7}}-{{x}^{2}}\] have?
Answer
446.1k+ views
Hint: First understand the meaning of the term ‘point of inflection’. Now, double differentiate the function \[f\left( x \right)\] to find the function \[f''\left( x \right)\]. Substitute \[f''\left( x \right)\] equal to 0 and find the values of x. These values of x obtained will be the points of inflection and our answer.
Complete step-by-step solution:
Here, we have been provided with the function \[f\left( x \right)={{x}^{7}}-{{x}^{2}}\] and we have been asked to determine the points of inflection for this function. But first we need to understand the meaning of the term ‘point of inflection’.
Now, in differential calculus, the point of inflection or inflection point is a point on a smooth curve at which the curvature sign changes. If we will consider the graph of a function then we can say that the point of inflection is a point where the function changes from being concave to convex or from being convex to concave. For a double differentiable function, to find the point of inflection we use the condition \[f''\left( x \right)=0\], where \[f\left( x \right)\] is the given function.
Let us come to the question. We have the function \[f\left( x \right)={{x}^{7}}-{{x}^{2}}\]. So, differentiating both the sides with respect to x, we get,
\[\Rightarrow f'\left( x \right)=7{{x}^{6}}-2x\]
Again, differentiating both the sides with respect to x, we get,
\[\Rightarrow f''\left( x \right)=42{{x}^{5}}-2\]
Substituting \[f''\left( x \right)=0\], we get,
\[\begin{align}
& \Rightarrow 42{{x}^{5}}-2=0 \\
& \Rightarrow 42{{x}^{5}}=2 \\
& \Rightarrow {{x}^{5}}=\dfrac{2}{42} \\
\end{align}\]
Cancelling the common factors, we get,
\[\Rightarrow {{x}^{5}}=\dfrac{1}{21}\]
Taking fifth root both the sides, we get,
\[\Rightarrow x={{\left( \dfrac{1}{21} \right)}^{\dfrac{1}{5}}}\]
Hence, \[x=\sqrt[5]{\dfrac{1}{21}}\] is the point of inflection, so there is only one point of inflection for the function \[f\left( x \right)\].
Note: One may note that \[f''\left( x \right)=0\] is not a sufficient condition for having a point of inflection. You must know about the points like: - stationary point, saddle point, critical point in differential calculus. There are many functions for which you will get \[f''\left( x \right)=0\] at point x = 0 but it will not be an inflection point. These things will be read in higher mathematics.
Complete step-by-step solution:
Here, we have been provided with the function \[f\left( x \right)={{x}^{7}}-{{x}^{2}}\] and we have been asked to determine the points of inflection for this function. But first we need to understand the meaning of the term ‘point of inflection’.
Now, in differential calculus, the point of inflection or inflection point is a point on a smooth curve at which the curvature sign changes. If we will consider the graph of a function then we can say that the point of inflection is a point where the function changes from being concave to convex or from being convex to concave. For a double differentiable function, to find the point of inflection we use the condition \[f''\left( x \right)=0\], where \[f\left( x \right)\] is the given function.
Let us come to the question. We have the function \[f\left( x \right)={{x}^{7}}-{{x}^{2}}\]. So, differentiating both the sides with respect to x, we get,
\[\Rightarrow f'\left( x \right)=7{{x}^{6}}-2x\]
Again, differentiating both the sides with respect to x, we get,
\[\Rightarrow f''\left( x \right)=42{{x}^{5}}-2\]
Substituting \[f''\left( x \right)=0\], we get,
\[\begin{align}
& \Rightarrow 42{{x}^{5}}-2=0 \\
& \Rightarrow 42{{x}^{5}}=2 \\
& \Rightarrow {{x}^{5}}=\dfrac{2}{42} \\
\end{align}\]
Cancelling the common factors, we get,
\[\Rightarrow {{x}^{5}}=\dfrac{1}{21}\]
Taking fifth root both the sides, we get,
\[\Rightarrow x={{\left( \dfrac{1}{21} \right)}^{\dfrac{1}{5}}}\]
Hence, \[x=\sqrt[5]{\dfrac{1}{21}}\] is the point of inflection, so there is only one point of inflection for the function \[f\left( x \right)\].
Note: One may note that \[f''\left( x \right)=0\] is not a sufficient condition for having a point of inflection. You must know about the points like: - stationary point, saddle point, critical point in differential calculus. There are many functions for which you will get \[f''\left( x \right)=0\] at point x = 0 but it will not be an inflection point. These things will be read in higher mathematics.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

What is a transformer Explain the principle construction class 12 physics CBSE

What are the major means of transport Explain each class 12 social science CBSE
