You need to solve the equation `f''(x) = 0` to find where the graph of function is concave up or down and what the inflection points are. You need to find the first derivative first such that:

`f'(x) = sqrt(x^2+4x+13) + 2x(x+2)/(2sqrt(x^2+4x+13)) + 2*2(x+2)/(2sqrt(x^2+4x+13))` `f'(x) = sqrt(x^2+4x+13) + x(x+2)/(sqrt(x^2+4x+13)) +...

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You need to solve the equation `f''(x) = 0` to find where the graph of function is concave up or down and what the inflection points are. You need to find the first derivative first such that:

`f'(x) = sqrt(x^2+4x+13) + 2x(x+2)/(2sqrt(x^2+4x+13)) + 2*2(x+2)/(2sqrt(x^2+4x+13))` `f'(x) = sqrt(x^2+4x+13) + x(x+2)/(sqrt(x^2+4x+13)) + 2(x+2)/(sqrt(x^2+4x+13))`

`f'(x) = (x^2 + 4x + 13 + 2x + x^2 + 2x + 4)/sqrt(x^2+4x+13)`

`f'(x) = (2x^2 + 8x + 17)/sqrt(x^2+4x+13)`

You need to find the second derivative using quotient law such that:

`f''(x) = ((2x^2 + 8x + 17)'*sqrt(x^2+4x+13) - (2x^2 + 8x + 17)*(sqrt(x^2+4x+13))')/(x^2+4x+13)`

`f''(x) = ((4x+8)*sqrt(x^2+4x+13) - (2x^2 + 8x + 17)*(x+2)/(sqrt(x^2+4x+13)))/(x^2+4x+13)`

`f''(x) = ((x+2)*(4sqrt(x^2+4x+13) - (2x^2 + 8x + 17)/(sqrt(x^2+4x+13))))/(x^2+4x+13)`

You need to solve `f''(x)=0` such that:

`(x+2)*(4sqrt(x^2+4x+13) - (2x^2 + 8x + 17)/(sqrt(x^2+4x+13))) = 0`

`x_1 = -2`

`4sqrt(x^2+4x+13) - (2x^2 + 8x + 17)/(sqrt(x^2+4x+13)) = 0`

Since `(sqrt(x^2+4x+13)) != 0,` then `4(x^2+4x+13) - (2x^2 + 8x + 17) = 0` such that:

`4x^2 + 16x + 52 - 2x^2 - 8x - 17 = 0`

Collecting like terms yields:

`2x^2 + 8x + 35 = 0`

You need to use quadratic formula such that:

`x_(1,2) = (-8+-sqrt(64 - 140))/4`

**Notice that the equation has no real roots, hence the function `f(x) ` has no inflection points over `[-6,5]` and `f''(x) gt 0` over `[-6,5], ` hence the graph of the function is concave up over `[-6,5].` **