Question

If \[f(x) = {x^2} - 3x\] , then the points at which \[f(x) = f'(x)\] are
A. \[1,3\]
B. \[1, - 3\]
C. \[ - 1,3\]
D. None of these

Answer 1

Hint: We are given a function and an equation. To find the points at which \[f'(x) = f(x)\] , we will find the first derivative of the function \[f(x)\] , that is, we will differentiate \[f(x)\] with respect to \[x\] and then put the value of \[f'(x)\] and \[f(x)\] in the given equality.

Formula used:
The following formulas are useful for solving the question
\[\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}\]
\[\begin{array}{l}a{x^2} + bx + c = 0\\ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\end{array}\]

Complete step by step answer:
We are given \[f(x) = {x^2} - 3x\]
Firstly, we will differentiate both sides with respect to \[x\]
\[\begin{array}{l}f'(x) = \dfrac{{d({x^2} - 3x)}}{{dx}}\\f'(x) = \dfrac{{d{x^2}}}{{dx}} - \dfrac{{d3x}}{{dx}}\\f'(x) = 2x - 3\end{array}\]
Now, we will find the points at which \[f(x) = f'(x)\] , that is,
\[{x^2} - 3x = 2x - 3\]
Here we will Simplify the above equation
\[\begin{array}{l}{x^2} - 5x + 3 = 0\\x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\x = \dfrac{{ - ( - 5) \pm \sqrt {{{( - 5)}^2} - 4 \times 1 \times 3} }}{{2(1)}}\\x = \dfrac{{5 \pm \sqrt {25 - 12} }}{2}\\x = \dfrac{{5 \pm \sqrt {13} }}{2}\end{array}\]
Therefore, the option (D) is correct.

Additional Information: An output variable and one or more input variables are the minimum number of variables in a function. An equation can have any number of variables and declares that two expressions are equal (none, one, or more). Although not all equations are functions, a function is frequently expressed as an equation. The derivative of a function of a real variable measures the change of the function value with respect to a change in its argument. Derivatives are a fundamental tool of calculus. It is also used to calculate the tangent of function.

Note: \[f(x) = {x^2} - 3x\] is a function where \[f(x)\] can also be written as \[y\] . Here \[y\] is the output variable and \[x\] is the input variable, that is, the value of \[y\] depends on the value of \[x\] . While \[f'(x) = f(x)\] is an equation as it suggests an equality between two different functions.