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Question 813683: State the number of real zeros. Approximate each real zero to the nearest tenth. Approximate the relative minima and relative maxima to the nearest tenth. : f(x)= -x^3+x^2-1
What I tried:
I successfully graphed the function.According to Descartes' Rule of Signs, there are 2 or zero positive real zeros and 1 negative real zero. Referring to the Intermediate Value Theorem, I decided that there was a real zero between [-1,1]. I did this:
f(-1)=1 -> f(-.9)=.539 -> f(-.8)=.152 -> f(-.7)=-.167.
Since there was a change in signs between -.8 and -.7, I got that the negative real zero is -.8.
My Question:
How do I find the positive real zeros, if there are any and how do you find the relative minima and relative maxima?
Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website! one real zero but it is negative.
x = -0.75488
two complex roots
x = 0.87744-0.74486 i
x = 0.87744+0.74486 i
max{-x^3+x^2-1} = -23/27 at x = 2/3
min{-x^3+x^2-1} = -1 at x = 0
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