Question

If x is real then minimum value of x^2-8x+17 is

Answer 1

$\frac{d {}^{2} }{dx {}^{2} } (f(x)) > 0$
2x-8
x=4
2>0
minimum=2

Answer 2

Answer :

The minimum value is 1 at x=4.

Explanation:

Given : Equation $x^2-8x+17$

To find : The minimum and maximum values of the equation ?

Solution :

To find the minimum points we find the first derivative and get the critical points then substitute that points in the second derivative.

Let $y=x^2-8x+17$

Derivate w.r.t x,

$y'=2x-8$

For critical points put y'=0,

$2x-8=0$

$x=4$

The second derivative is

$y''=2$

The second derivative is greater than zero.

At x=4,

$(4)^2-8(4)+17=16-32+17=1$

The minimum value is 1 at x=4.