Question 10 on Precalculus

Question

f(x)=x^3+3x^2-9x+5

Math problem

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Answer 1

f(x) = x^3 +3x^2 -9x + 5

is a cubic function, degree 3

Its y intercept, when graphed is 5, the constant term

there is no global maximum or minimum. It's end behavior approaches infinity and negative infinity

but it has a local or relative max and min, found by solving the derivative set =0

f'(x) = 3x^2 +6x -9 = 0

x^2 +2x -3 = 0

factor to get

(x+3)(x-1) = 0

set each factor =0 and solve for x

x=1 and -3.

-3 is the relative maximum,

1 is the relative minimum

or the points (1,0) and (-3, 32)

Using Descartes' Rule of Signs (count the sign changes of f(x), then count sign changes of f(-x))

there are two potential positive roots or x intercepts

there is one potential negative root or x intercept

the x intercepts, if any real ones, would be each side of the relative extrema

to find them set f(x) = 0 and solve for x, but that's not generally easy to do, with higher than 2nd degree equations. But you may notice x=1 is one solution, then divide the 3rd degree polynomial by x-1 to get a quadratic which you could easily solve for 2 other solutions x^2+4x-5 is the quotient when you divide f(x) by x-1. the quadratic factors into (x+5)(x-1). the other root is -5

The other possible real roots are one positive number and one negative number. Since x=1 is root and also a relative minimum, that means the root has multiplicity 2, and graphically the curve is tangent to the x axis at x=1, or the point (1,0). That leave only one other possible real root a negative number, less than -3, the relative maximum. It exists, it's x=-5, where the curve crosses the x-axis

in one sense there are 3 roots, 1, 1, and -5, with one root repeating itself. Or they usually say it's just 2 roots, where one has multiplicity of two.

That covers the usual questions that arise with a math problem involving a polynomial function. With the above information, you should be able to sketch a rought graph of the function. It has a "U" shapre on the right side, and an inverted U shape on the left side. It approaches infinity on the right side, and negative infinity on the left side.

There is an inflection point where the U and inverted U shapes are connected. It's where f"(x) = 0, (-1,16), where the concavity changes, from concave down to concave up f'(x) = 3x^2 +6x -9. f"(x) = 6x+ 6 =0, x=-1= the x coordinate of the inflection point. Plug it into f(x) to get

f(-1) = (-1)^3 +3(-1)^2-9(-1) +5 = -1+3+9+5 =16=the y coordinate of the inflection point. it's horizontally half way between x=-3 and x=1 the x coordinates of the local extrema. the y coordinate is midway vertically from the y coordinates of the local extrema. The inflection point is midway, exactly half way between the two local extrema.

.From left side to the local maximum the slope of the curve is positive, from the local minumum to the right the slope is positive. Between the local max to local min, the slope is negative

From the left side to the inflection point the slope is decreasing. from the inflection point to the right the slope is increasing