Question

can someone please help me with this question What is the minimum value of the quadratic function f(x) = x² - 2x + 7

Answer 1

minimum value of the quadratic function f(x) = x² - 2x + 7 is at x=1 & is (1, 6).

Step-by-step explanation:

Here we have ,  f(x) = x² - 2x + 7 or $f(x) = x^2 - 2x + 7$ . We need to find the minimum value of f(x) for which we need to differentiate it one time and equate it to zero . Value of x at which first differentiation of f(x) is zero will be the minimum value of function  . Let's solve this:

$f(x) = x^2 - 2x + 7$

⇒ $f(x) = x^2 - 2x + 7$

⇒ $\frac{df(x)}{dx} = \frac{d(x^2 - 2x + 7)}{dx}$

⇒ $\frac{df(x)}{dx} = \frac{d(x^2)}{dx} - \frac{d(2x)}{dx} + \frac{d(7)}{dx}$

⇒ $\frac{df(x)}{dx} = 2x-2 = 0$

⇒ $2x-2 = 0$

⇒ $x =1$

Now, value of function at x=1 is :

$f(x) = x^2 - 2x + 7$

⇒ $f(x) = x^2 - 2x + 7$

⇒ $f(1) = 1^2 - 2(1) + 7$

⇒ $f(1) = 8- 2$

⇒ $f(1) = 6$

Therefore, minimum value of the quadratic function f(x) = x² - 2x + 7 is at x=1 & is (1, 6).