Question

Find the absolute maximum and absolute minimum values of f on the given interval. f(x)=ln(x^2+x+1), [-1,1]

Answer 1

well you can differentiate and find the stationary point(s)... I think the stationary point is a min... and text the endpoints for max values...

Answer 2

idk how to work it out completely

Answer 3

ok... so the end point

substitute x = -1 and x = 1 into the equation...

then the derivative is

\[\frac{dy}{dx} = \frac{2x + 1}{x^2 + x + 1}\]

so the stationary point occurs when

2x + 1 = 0

hope this helps

Answer 4

The answer in the back of the book says f(1)=ln3, f(-1/2)= ln 3/4... don't understand

Answer 5

so a stationary point occurs at x = -1/2


test at x = -1/4 and x = -3/4 for a change in the slope of the tangent...

x = -1/4 dy/dx > 0 x = -3/4 dy/dx < 0

so x = -1/2 is a minimum substitute it to find the value

hope this makes sense

Answer 6

f(1) = ln(3) is the max value of the curve between [-1, 1]

f(-1/2) = ln(3/4) is the minimum value of the curve in the same domain [-1,1]

Answer 7

the cuve looks like