Find f(x) given x-axis and lowpoint??
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What does that mean?I have to find f(x) given x-axis (-3,0) (0,0)
and the low point -2.25;-8.5
and this photo:
Can you help me?
What are your thoughts?
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Otis
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Your reply to Subhotosh's question may be pending approval (I don't see any reply), but I'll post anyway.
I'm guessing that you meant to say, "given x-intercepts (-3,0) (0,0)".
One way to find function f is to write and solve a system of equations. Did you solve systems of equations in precalc?
The given chart shows information about the shape of the graph. Remember that f' is the slope of the graph of f, so the sign of f' tells us where the graph of function f is increasing (positive slope) or decreasing (negative slope). Where f' is zero, the graph of f is at a local minimum or maximum.
The sign of f'' tells us about the graph's concavity. Where f'' is positive, the graph of f is concave up, and the graph of f is concave down where f'' is negative.
Using this information, you can sketch a rough graph of function f, from x=-3 to x=0.
If you studied the graphs of polynomial equations in precalc, then you'll hopefully recognize that a fourth-degree polynomial generates a curve with that shape.
Using symbol y instead of f(x), function f is defined symbolically as:
y = a*x^4 + b*x^3 + c*x^2 + d*x + e
The graph goes through the origin. The only way that can happen is with e=0.
Therefore, y = a*x^4 + b*x^3 + c*x^2 + d*x
Now, you determine the first and second derivatives (y' and y'').
Substituting known values for x, y, y', and y'' yields a system of four equations to solve, for the coefficients a,b,c,d.
An intercept: y = 0 when x = -3
The local minimum: y = -8.5 when x =-2.25
The derivatives:
y' = 0 when x = ? (look it up on the chart)
y'' = 0 when x = ? (look it up on the chart)
If you've forgotten too much precalc, let me know, and I'll guide you through the steps one at a time.
Please start by calculating y' and y''. Show us what you get.
I'm guessing that you meant to say, "given x-intercepts (-3,0) (0,0)".
One way to find function f is to write and solve a system of equations. Did you solve systems of equations in precalc?
The given chart shows information about the shape of the graph. Remember that f' is the slope of the graph of f, so the sign of f' tells us where the graph of function f is increasing (positive slope) or decreasing (negative slope). Where f' is zero, the graph of f is at a local minimum or maximum.
The sign of f'' tells us about the graph's concavity. Where f'' is positive, the graph of f is concave up, and the graph of f is concave down where f'' is negative.
Using this information, you can sketch a rough graph of function f, from x=-3 to x=0.
If you studied the graphs of polynomial equations in precalc, then you'll hopefully recognize that a fourth-degree polynomial generates a curve with that shape.
Using symbol y instead of f(x), function f is defined symbolically as:
y = a*x^4 + b*x^3 + c*x^2 + d*x + e
The graph goes through the origin. The only way that can happen is with e=0.
Therefore, y = a*x^4 + b*x^3 + c*x^2 + d*x
Now, you determine the first and second derivatives (y' and y'').
Substituting known values for x, y, y', and y'' yields a system of four equations to solve, for the coefficients a,b,c,d.
An intercept: y = 0 when x = -3
The local minimum: y = -8.5 when x =-2.25
The derivatives:
y' = 0 when x = ? (look it up on the chart)
y'' = 0 when x = ? (look it up on the chart)
If you've forgotten too much precalc, let me know, and I'll guide you through the steps one at a time.
Please start by calculating y' and y''. Show us what you get.
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Your reply to Subhotosh's question may be pending approval (I don't see any reply), but I'll post anyway.
I'm guessing that you meant to say, "given x-intercepts (-3,0) (0,0)".
One way to find function f is to write and solve a system of equations. Did you solve systems of equations in precalc?
The given chart shows information about the shape of the graph. Remember that f' is the slope of the graph of f, so the sign of f' tells us where the graph of function f is increasing (positive slope) or decreasing (negative slope). Where f' is zero, the graph of f is at a local minimum or maximum.
The sign of f'' tells us about the graph's concavity. Where f'' is positive, the graph of f is concave up, and the graph of f is concave down where f'' is negative.
Using this information, you can sketch a rough graph of function f, from x=-3 to x=0.
If you studied the graphs of polynomial equations in precalc, then you'll hopefully recognize that a fourth-degree polynomial generates a curve with that shape.
Using symbol y instead of f(x), function f is defined symbolically as:
y = a*x^4 + b*x^3 + c*x^2 + d*x + e
The graph goes through the origin. The only way that can happen is with e=0.
Therefore, y = a*x^4 + b*x^3 + c*x^2 + d*x
Now, you determine the first and second derivatives (y' and y'').
Substituting known values for x, y, y', and y'' yields a system of four equations to solve, for the coefficients a,b,c,d.
An intercept: y = 0 when x = -3
The local minimum: y = -8.5 when x =-2.25
The derivatives:
y' = 0 when x = ? (look it up on the chart)
y'' = 0 when x = ? (look it up on the chart)
If you've forgotten too much precalc, let me know, and I'll guide you through the steps one at a time.
Please start by calculating y' and y''. Show us what you get.
Thank you for the help, if you can help me with the next step that would be great. Unfortunately I'm starting school again after many years, so I have forgotten about that.
Otis
Elite Member
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- Apr 22, 2015
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If you're saying that you've forgotten your precalculus material, then you may need to consider dropping calculus and taking a refresher course in precalc (and, perhaps, algebra and trigonometry, too).
Calculus is a heady subject. I have worked with many math students who try to pick up where they left off, after not having done math for a long time. It never goes well because math instruction generally builds upon previous material. Some students think that they can refresh their memory on algebra, trigonometry, and precalculus topics while they try to learn calculus at the same time. That's okay, if you're very self-motivated, have lots of available time, and have only forgotten relatively few things. For those students who have forgotten almost everything, I have never seen such a plan succeed.
I don't know how much you've forgotten.
If you want to try relearning the prerequisites while also studying calculus, then you will definitely need a face-to-face tutor. We cannot teach prerequisite classroom material, on these boards. If you already feel completely lost, I would encourage you to speak with your instructor or adviser, to see what they suggest. Ask about tutoring resources, too.
Be prepared to devote up to two dozen hours each week, working outside of class.
If you would like to try finding function f by setting up and solving a system of equations, then the next step is for you to find the first and second derivatives of:
a*x^4 + b*x^3 + c*x^2 + d*x
You can use the power rule, to determine these derivatives. Please try, and post your work. :cool:
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Otis
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Your original posts says that you're looking for f(x), but I just noticed that the given chart uses the name m(x), instead. Whether we call the function f or m doesn't matter to me, but it appears that the exercise uses the name m. I'll continue using the name f, in my replies.
The graph of function f is not a parabola. We can tell this by examining the given sign chart for the second derivative. The second derivative values are positive in some intervals, and negative elsewhere. This means that the graph of f changes concavity.
A parabola never changes concavity (it's always completely concave up or completely concave down, not a mixture of both). Therefore, f(x) cannot be a quadratic polynomial (i.e., second-degree).
It can't be a cubic polynomial (i.e., third-degree), either. Your graph has a local minimum; cubic function graphs don't have local minimums or maximums.
Again, from precalc material, the polynomial must be of degree four (with no constant term, as explained earlier).
f(x) = a*x^4 + b*x^3 + c*x^2 + d*x
Your eventual goal is to find the values for the coefficients a,b,c,d.
That's a misstatement because the x-axis is not a root. It is a horizontal line that extends forever in each direction, while roots, on the other hand, are individual values of x which cause y to equal zero. Roots appear on a graph as single points, not a line; we call these points x-intercepts because it's at these locations that a graph crosses (or touches) the x-axis.
That's why the names "root" and "x-axis" do not mean the same thing.
It is given that the x-intercepts are at (-3,0) and (0,0), so we can say that the roots are -3 and 0. (Function f has no positive root.)
I'll wait to see your efforts on differentiating a*x^4 + b*x^3 + c*x^2 + d*x. Cheers
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Otis
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Here's a graph of function f, in the vicinity of the intervals provided by the given sign chart.
Look at the concavity between x=-1.5 (that's the decimal form of -3/2) and the origin. It's concave down, yes? That agrees with the negative sign of the second derivative (see the chart) in the same interval.
It's concave up elsewhere, as shown by the positive signs in the chart. :cool:
Look at the concavity between x=-1.5 (that's the decimal form of -3/2) and the origin. It's concave down, yes? That agrees with the negative sign of the second derivative (see the chart) in the same interval.
It's concave up elsewhere, as shown by the positive signs in the chart. :cool:
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