# How do I find the critical points of x^2-7x+ 13)2^x, and how can I apply the second derivative test to this? You need to find the critical points of the given function, hence, you need to differentiate the function, using the product rule, such that:

`(f(x)*g(x))' = f'(x)*g(x) + f(x)*g'(x)`

Identifying `f(x) = x^2 - 7x + 13` and `g(x) = 2^x` , yields:

`f'(x) = (x^2 - 7x + 13)' => f'(x) = 2x - 7`

`g'(x) = 2^x*ln 2`

`(f(x)*g(x))' = (2x - 7)*2^x - (x^2 - 7x + 13)*(2^x*ln 2)`

Factoring out `2^x` , yields:

`(f(x)*g(x))' = 2^x(2x - 7 - x^2*ln2 + 7x*ln 2 - 13ln 2)`

`(f(x)*g(x))' = 2^x(- x^2*ln2 + x(2 + 7ln 2) - 13ln 2 - 7)`

You need to solve for x the following equation to evaluate the critical points of the function, such that:

`(f(x)*g(x))' = 0 => 2^x(- x^2*ln2 + x(2 + 7ln 2) - 13ln 2 - 7) = 0`

Since `2^x > 0` yields `- x^2*ln2 + x(2 + 7ln 2) - 13ln 2 - 7 = 0.`

`x^2*ln2 - x(2 + 7ln 2) + 13ln 2 + 7 = 0`

Using quadratic formula, yields:

`x_(1,2) = (2 + 7ln 2 +- sqrt(4 + 28ln 2 + 49(ln^2 2) - 4ln2(13ln 2 + 7)))/(ln 4)`

`x_(1,2) = (2 + 7ln 2 +- sqrt(4 + 28ln 2 + 49(ln^2 2) - 52(ln^2 2) - 28ln 2))/(ln 4)`

`x_(1,2) = (2 + 7ln 2 +- sqrt(4 - 3ln^2 2))/(ln 4)`

`x_(1,2) ~~ (2 + 7ln 2 +- 1.6)/(1.386)`

`x_1 ~~ 6.097; x_2 ~~ 3.788`

Hence, evaluating the critical values of the function yields `x_1 ~~ 6.097; x_2 ~~ 3.788.`

The function will reach the maximum point at `x_2 ~~ 3.788 ` and it will reach its minimum at `x_1 ~~ 6.097.`

The second derivative will be evaluated using again the product rule, such that:

(f(x)*g(x))'' = 2^x*ln 2(- x^2*ln2 + x(2 + 7ln 2) - 13ln 2 - 7) - 2^x(- 2x*ln2 + (2 + 7ln 2))

Approved by eNotes Editorial Team