A quadratic equation is shown. (x + 6)(x + 2) = x
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The vertex (minimum) of the quadratic ax² +bx +c is located at x=-b/(2a). This means the minimum value of f(x) will be found at x = -3/(2*1) = -1.5.
Since the vertex of the quadratic is less than 0, the maximum value of the quadratic will be found at x=2, the end of the interval farthest from the vertex.
On the given interval, ...
the absolute minimum value of f is f(-1.5) = ln(1.75) ≈ 0.559616
the absolute maximum value of f is f(2) = ln(14) ≈ 2.639057
Since the vertex of the quadratic is less than 0, the maximum value of the quadratic will be found at x=2, the end of the interval farthest from the vertex.
On the given interval, ...
the absolute minimum value of f is f(-1.5) = ln(1.75) ≈ 0.559616
the absolute maximum value of f is f(2) = ln(14) ≈ 2.639057