Asked by Em
Determine the maximum possible number of turning points for the graph of the function.
f(x) = 8x^3 - 3x^2 + -8x - 22
-I got 2
f(x) = x^7 + 3x^8
-I got 7
g(x) = - x + 2
I got 0
How do I graph f(x) = 4x - x^3 - x^5?
f(x) = 8x^3 - 3x^2 + -8x - 22
-I got 2
f(x) = x^7 + 3x^8
-I got 7
g(x) = - x + 2
I got 0
How do I graph f(x) = 4x - x^3 - x^5?
Answers
Answered by
Reiny
do you know calculus???
first I factored it to
f(x) = -x(x^4 + x^2 - 1)
treating the big bracket as a quadratic, I found x=0, x = ±1.11 and 2 complex roots.
finding the second derivative, setting that equal to zero and solving I got x=0 and x=5/2
so there are two points of inflection, namely at x=0 and at x=5/2
Lastly since the highest power term was negative and an odd exponent, the curve "drops" into the fourth quadrant
so your graph comes down from the second quadrant, crosses at -1.11, comes back up crossing at 0, then comes back down to cross at 1.11. It does a little S bend at x=5/2
You will have to use a different scale for your x and y axes.
first I factored it to
f(x) = -x(x^4 + x^2 - 1)
treating the big bracket as a quadratic, I found x=0, x = ±1.11 and 2 complex roots.
finding the second derivative, setting that equal to zero and solving I got x=0 and x=5/2
so there are two points of inflection, namely at x=0 and at x=5/2
Lastly since the highest power term was negative and an odd exponent, the curve "drops" into the fourth quadrant
so your graph comes down from the second quadrant, crosses at -1.11, comes back up crossing at 0, then comes back down to cross at 1.11. It does a little S bend at x=5/2
You will have to use a different scale for your x and y axes.
Answered by
Reiny
BTW, your first two answers are correct
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