Question

the polynomial function $f(x)=x^{2}-4x + 1$ has a critical point at which of the following x - values?
a. $x = 2$
b. $x = 1$
c. $x = 0$
d. $x=-2$

Explanation:

Step1: Find the derivative

The derivative of $F(x)=x^{2}-4x + 1$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ is $F^\prime(x)=2x-4$.

Step2: Set derivative equal to zero

To find the critical points, we set $F^\prime(x) = 0$. So, $2x-4=0$.

Step3: Solve for x

Add 4 to both sides: $2x=4$. Then divide both sides by 2, we get $x = 2$.

Answer:

A. $x = 2$