Find the absolute minimum and absolute maximum values of f on the given interval. f(x) = ln(x2 + 3x + 4), [−2, 2]
1 answer:
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
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Answer and Step-by-step explanation:
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In the equation, the m stands for the slope in the equation, and the b stands for the y intercept. Slope can also be found by taking the rise/run of a graph.
So, the slope of this line is 25, and the y-intercept 100.
I have also included the this equation on a graph as well.
I hope that this helps.
Answer:
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