Question

If $y = 3{x^2} - 2x + 1$, find the value of x for which ${y^1} = 0$.

Answer 1

Hint: In this question, we have given a function. First, we have to differentiate that function with respect to $x$. Then, we will evaluate the differentiated term to zero. Now, we can find the value of $x$ from the differentiated equation.

Complete step by step answer:
In the above question, it is given that $y = 3{x^2} - 2x + 1$.
Now, we will differentiate the above function with respect to x.
On differentiation, we get
${y^1} = 6x - 2$
Now, we have given a condition that ${y^1} = 0$.
Therefore, on putting the value of ${y^1}$, we get
$6x - 2 = 0$
$ \Rightarrow 6x = 2$
$ \Rightarrow x = \dfrac{2}{6}$
$ \therefore x = \dfrac{1}{3}$

Therefore, the required value of x is $\dfrac{1}{3}$.

Note: In the above question, if is given instead of ${y^1}$, then we have to differentiate the given function twice and then we will use the given condition to find the value of $x$. Here, ${y^1}$ is given as zero. Therefore, we can say that the value of x we will get here will be the point of either maxima or minima of the given function. If we want to find whether it is a point of maxima or minima, we have to differentiate it one more time. Now, if we put the value of $x$ in the double differentiated term and if we get a negative number, we can say that it is a point of maxima. If it comes out to be positive, we can say that it is a point of minima and if it is zero then it is the point of inflexion.