Solve Tangent of Line Homework | y=f(x) Slope 4√2x+7

In summary, the function f(x) is equal to ((4(2x+7)^3/2) / (3)) - 52, where the graph of y=f(x) passes through the point (9/2, 100/3) and the tangent line to the graph at any point (x,y) has the slope 4*sqrt(2x+7). The constant of integration, +C, must be included in the function in order for it to satisfy the given conditions.
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Homework Statement


The graph of y = f(x) passes through the point (9/2, 100/3). Also the tangent line to the graph at any point (x,y) has the slope 4*sqrt(2x+7). Find f(x)

Homework Equations


The Attempt at a Solution


I am very lost with this as I can't find much info in my textbook. Any help where to start? I am assuming I am trying to find the equation of the line y=mx+c given the slope of the tangent and current points.
 
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  • #2


menco said:

Homework Statement


The graph of y = f(x) passes through the point (9/2, 100/3). Also the tangent line to the graph at any point (x,y) has the slope 4*sqrt(2x+7). Find f(x)


Homework Equations





The Attempt at a Solution


I am very lost with this as I can't find much info in my textbook. Any help where to start?

Given a function [itex]y=f(x)[/itex], how does one find the slope of the tangent line? (If you aren't sure, you had better open up your textbook and find out the definition of tangent line)

I am assuming I am trying to find the equation of the line y=mx+c given the slope of the tangent and current points.

No, you are asked to find the original function [itex]f(x)[/itex] (which will not be a straight line), not the tangent line at some point.
 
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I know how to find the slope and equation of a tangent line fairly easy but trying to reverse it is confusing me
 
  • #4


menco said:
I know how to find the slope and equation of a tangent line fairly easy

Again, describe how to find the slope of the tangent line to a function [itex]y=f(x)[/itex]. (Don't say that you know how, demonstrate that you know)
 
  • #5


To find the slope of a tangent line take the derivative of the function and substitute in the point of contact if known.
 
  • #6


menco said:
To find the slope of a tangent line take the derivative of the function and substitute in the point of contact if known.

Right, and so at a general point [itex]x[/itex], the slope is just [itex]f'(x)[/itex]. So, what can you say about [itex]f(x)[/itex] if the slope of the tangent line at a point [itex]x[/itex] is [itex]4\sqrt{2x + 7}[/itex]?
 
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the rate of change of the function is 4*sqrt(2x+7)?
 
  • #8


menco said:
the rate of change of the function is 4*sqrt(2x+7)?

Yes, [itex]f'(x)=4\sqrt{2x+7}[/itex].

So, [itex]f(x)=?[/itex]...
 
  • #9


The opposite of the derivative is the anti-derivative, also called the "indefinite integral". Have you studied those?
 
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  • #10


Yes we have just started integrals, by using substitution

I found f(x) = (4(2x+7)^3/2) / 3

Does the point (9/2, 100/3) have anything to do with the problem?
 
  • #11


Yes, the graph of y=f(x) passes through that point, so you should have that 100/3 = f(9/2). Is that the case for your f(x)? If not, how do you fix it?
 
  • #12
f(9/2) = 256/3

So it is not the case, I'm a little unsure of what you mean by fix it?
 
  • #13
##\frac{4}{3}(2x+7)^{3/2}## is not the only function whose derivative is ##4\sqrt{2x+7}##. You need to find another one, one where f(9/2)=100/3.
 
  • #14
menco, take this equation for example: [itex]\int x^2 = \frac{x^3}{3} + C[/itex], do you remember why we put +C there?
 
  • #15
Ah yes i see I forgot all about +C, which is the constant of integration.

So if I use 100/3 = 256/3 + C, C = (-52)

Therefore the final function will be ((4(2x+7)^3/2) / (3)) - 52

then when f(9/2) = 100/3
 

1. What is the slope of the line y=f(x) with the equation 4√2x+7?

The slope of a line is determined by the coefficient of x in its equation when written in slope-intercept form, y=mx+b. In this case, the coefficient of x is 4√2, so the slope is 4√2.

2. How do I solve for the tangent of a line with a given equation?

To solve for the tangent of a line, you will need to use the derivative of the function. In this case, the derivative of y=f(x) is y'=4√2. The tangent of a line at a given point is equal to the slope of the line at that point, so the slope of the line at any point on the line y=f(x) is 4√2.

3. Can I use the tangent of a line to find the equation of a line?

Yes, the slope of a line (which is equal to the tangent of the line) can be used to find the equation of a line. The general equation of a line is y=mx+b, where m is the slope and b is the y-intercept. By plugging in the slope (4√2) and a point on the line (such as (0,7)), you can solve for the y-intercept and therefore find the equation of the line.

4. How does the tangent of a line relate to the concept of derivatives?

The derivative of a function is the slope of the tangent line at any given point on the function. In other words, the derivative of a function gives us the slope of the tangent line at any point on the function. In this case, the derivative of y=f(x) is y'=4√2, so the slope of the tangent line at any point on the line y=f(x) is 4√2.

5. Is there a way to graph the tangent of a line?

Yes, you can graph the tangent of a line by plotting points on the line using the slope (4√2) and a given point on the line (such as (0,7)). You can also use the derivative of the function to find the equation of the tangent line and plot it on the same graph as the original function.

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