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 [tex]f(x) = x^2 - 2x + 7[/tex] . 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:
[tex]f(x) = x^2 - 2x + 7[/tex]
⇒ [tex]f(x) = x^2 - 2x + 7[/tex]
⇒ [tex]\frac{df(x)}{dx} = \frac{d(x^2 - 2x + 7)}{dx}[/tex]
⇒ [tex]\frac{df(x)}{dx} = \frac{d(x^2)}{dx} - \frac{d(2x)}{dx} + \frac{d(7)}{dx}[/tex]
⇒ [tex]\frac{df(x)}{dx} = 2x-2 = 0[/tex]
⇒ [tex]2x-2 = 0[/tex]
⇒ [tex]x =1[/tex]
Now, value of function at x=1 is :
[tex]f(x) = x^2 - 2x + 7[/tex]
⇒ [tex]f(x) = x^2 - 2x + 7[/tex]
⇒ [tex]f(1) = 1^2 - 2(1) + 7[/tex]
⇒ [tex]f(1) = 8- 2[/tex]
⇒ [tex]f(1) = 6[/tex]
Therefore, minimum value of the quadratic function f(x) = x² - 2x + 7 is at x=1 & is (1, 6).
