# 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?

### 1 Answer | Add Yours

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))**