(a) Find the slope of the tangent to the curve at the point where x = a.

In summary: I think.In summary, the equation of a line is slope, point. Homework Statement Following is a conversation. A student is trying to find the slope of a tangent line to a curve. Their teacher provides a summary of the content before the student outputs anything. The student provides a summary of the conversation. The student has trouble writing neatly, which is noted by their teacher. The student still can't output the slope of the line, and notes that all derivations can be linked back to the slope, point equation. In summary, the student is struggling with the slope of the line. The student's teacher provides a summary of the content before the student outputs anything. The
  • #1
Pondera
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Homework Statement


Consider the following curve.
y = 4 + 4x2 - 2x3.

(a) Find the slope of the tangent to the curve at the point where x = a.

Homework Equations


m=y(a+h) - y(a)/h

The Attempt at a Solution


1. m=y(a+h) - y(a)/h = -a^3-a^2+3ah+h^2-4
2. 6a^2
3. -6a^2+8a

I know the derivative is equal to the slope, but I don't know how to work that into this specific equation and point.

Thank you for any and all help.
 
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  • #2
Do you know any differentiation formulas, or do you have to do this by using the definition of the derivative?

The equation you give as the relevant equation works for equations of lines, but not for any other equations.

If you have to use the deriviative definition, let's call your function f(x) = -2x^3 + 4x^2 + 4.

By the definition,
[tex]f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}[/tex]

You skipped a lot of steps apparently in your attempt at a solution. Start by evaluating f(x + h), and subtract f(x) from that. Then divide each term in f(x + h) - f(x) by h, and finally, take the limit as h approaches zero.

That will be f'(x). To get f'(a), substitute a for x in your derivative function.
 
  • #3
This is the only formula we've learned so far, though I'm sure we'll be learning differentiation formulas shortly.

After taking f(x+h)-f(x)/h I came up with -6x^2-6xh-2h^2-4x^2/h-4/h, then plugging in 0 for h to take the lim h->0 I come up with -6x^2.

I have a couple more sections of this problem to work through before I can submit my answer to see if it's correct.

Thank you for your help so far!

Edit: My answer wasn't correct, so I'm going to try and work it out from the beginning again. Any suggestions? I think I see what I did (didn't work out the 4(x+h)^2 part properly), but we'll see.

Third attempt: -6x^2+8x, so -6a^2+8a
 
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  • #4
You need parentheses.
After taking f(x+h)-f(x)/h I came up with -6x^2-6xh-2h^2-4x^2/h-4/h,
First off, its (f(x + h) - f(x))/h, not f(x+h)-f(x)/h. These are different. And with your result, I'm not sure even how to interpret it.

Let me break it down for you:
1. Calculate f(x + h)

2. Calculate f(x + h) - f(x)

3. Calculate [f(x + h) - f(x)]/h

4. Take the limit, as h approaches 0, of [f(x + h) - f(x)]/h

In your reply, please provide the results you get from each step above.
 
  • #5
1. -2x^3+4x^2-6x^2h-6xh^2+8xh-2h^3+4h^2+4
2. -6x^2h-6xh^2+8xh-2h^3+4h^2
3. -6x^2-6xh+8x-2h^2+4h
4. -6x^2+8x
 
  • #6
Bingo and congratulations!
In the last line, what you have is f'(x).
What is f'(a)? That is the slope of the tangent line to the curve at x = a.

Some advice to make your work more readable on this forum -- use parentheses where they're needed, and space things out. You're more likely to get help if people don't have to work so hard to understand what you're trying to do.

1. -2x^3 + 4x^2 - 6hx^2 - 6xh^2 + 8xh - 2h^3 + 4h^2 + 4
2. -6hx^2 - 6xh^2 + 8xh - 2h^3 + 4h^2
3. -6x^2 -6xh + 8x - 2h^2 + 4h
4. -6x^2+8x

Notice that I moved h in the first two lines, since -6x^2h isn't as clear as -6hx^2, and makes the reader think that maybe h is in the exponent.
 
  • #7
Isn't f'(a) just -6a^2+8a?

Thank you again for the help and the suggestions on formatting!

Edit: Apparently it is, and I was able to get the other parts of the problem (finding the equation of the tangent line given a point) easily enough. I really can't thank you enough!
 
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  • #8
Not just formatting on this forum- your teacher will appreciate clearer writing too!

The tangent line you want has slope f'(a)= -6a^2+ 8a and passes through the point (a, f(a))= (a, 4 + 4a^2 - 2a^3). Do you know the "slope, point" formula for a line?
 
  • #9
HallsofIvy said:
Not just formatting on this forum- your teacher will appreciate clearer writing too!

The tangent line you want has slope f'(a)= -6a^2+ 8a and passes through the point (a, f(a))= (a, 4 + 4a^2 - 2a^3). Do you know the "slope, point" formula for a line?

Ah now I know why all my teachers hated me! :smile:

I still can't write neatly, it's something other people do. Lord praise the word processor.

That equation is the fundamental of differentiation, all of the various rules can be linked to it in one way or another. That's why you always learn it first, once you get why that works, then it makes the actual differentiation more readily and intuitively understandable.

That said god bless the rules eh. :smile: [itex]nx^{n-1}[/itex]
 

Related to (a) Find the slope of the tangent to the curve at the point where x = a.

1. What is the slope of the tangent to the curve at the point where x = a?

The slope of the tangent to the curve at the point where x = a is the instantaneous rate of change of the curve at that specific point. It can be calculated using the derivative of the curve at that point.

2. How is the slope of the tangent to the curve at the point where x = a calculated?

The slope of the tangent to the curve at the point where x = a is calculated using the derivative of the curve at that point. This is typically done using the limit definition of the derivative or by using differentiation rules.

3. Why is finding the slope of the tangent to the curve at the point where x = a important?

Finding the slope of the tangent to the curve at the point where x = a is important because it helps us understand the behavior of the curve at that specific point. It can also be used to find the equation of the tangent line, which can be useful in solving real-world problems.

4. Can the slope of the tangent to the curve at the point where x = a be negative?

Yes, the slope of the tangent to the curve at the point where x = a can be negative. The slope can be positive, negative, or zero, depending on the behavior of the curve at that specific point.

5. Is there a difference between the slope of the tangent to the curve and the slope of the curve at the point where x = a?

Yes, there is a difference between the slope of the tangent to the curve and the slope of the curve at the point where x = a. The slope of the tangent is the instantaneous rate of change at that specific point, while the slope of the curve is the average rate of change between two points on the curve.

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