Solving f(x), Slope & Tangent Line: Help Me Find Answers!

In summary, the conversation discusses finding the slope and tangent line for a given function, as well as finding the instantaneous rate of change for another function. The method of finding the derivative and substituting values is mentioned, but there is also a mention of using the gradient to find the slope. The conversation ends with the understanding that the derivative is necessary to find the instantaneous rate of change.
  • #1
VBoy336
7
0
can you guys help me how to do this problem?

f(x) = 3x^2 - 1 (3,8)
Find slope and tangent line,

I have answer 6, is that right? and can you guys show the step to get the answer, thanks =) (tangent line i already know how to do)

oh and this one,

y = x^2 - 4
Find instantaneous rate of change [3,5]

(if there was one point, i could do it, but i don't know how to do with 2 point,) can you show steps too,


thanks alot
 
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  • #2
For the first one, you find the derivative first.
[tex]f(x) = 3x^2-1 \ \longrightarrow \ f'(x) = 6x [/tex] then you substitude for x. That doesn't give 6. Now that you have the slope you can find the tangent line.

For the second line, I think that would be the gradient of the secant line through those points.
 
  • #3
for this one,
y = x^2 - 4
Find instantaneous rate of change [3,5]

i don't really get what you are saying, sorry,

so how do I solve it

urg
 
  • #4
VBoy336 said:
for this one,
y = x^2 - 4
Find instantaneous rate of change [3,5]

i don't really get what you are saying, sorry,

so how do I solve it

urg

@3, y = 5
@5, y = 11

Now you have 2 points, P1(3,5) & P2(5,11).
What is the gradient of the line that passes through these points ?
 
  • #5
@5 , do you mean 21 ?

and i don't think i learn gradient yet,

in class, when we finding instantaneous rate of change, we use limit x approaching a number, then solve it,

is there another way to find the instantaneous rate of change in this problem?
 
  • #6
yeah i mean 21, lol sry. Gradient = slope
 
  • #7
so the instant.. rate of change is 8 ?

aha
 

1. How do I solve for f(x)?

Solving for f(x) involves finding the value of the function at a specific input or x-value. This can be done by plugging in the given x-value into the function and simplifying the expression. For example, if f(x) = 2x + 3 and we want to find the value of f(5), we would plug in 5 for x and get f(5) = 2(5) + 3 = 13.

2. What is slope and how do I calculate it?

Slope is a measure of how steep a line is and is represented by the letter m. It can be calculated using the formula m = (y2 - y1) / (x2 - x1), where (x1, y1) and (x2, y2) are two points on the line. This formula is also known as the "rise over run" method, where the rise is the change in y-values and the run is the change in x-values.

3. How do I find the tangent line to a curve?

To find the tangent line to a curve at a specific point, you will need to find the slope of the curve at that point. This can be done by taking the derivative of the function and then plugging in the x-value of the point to find the slope. Once you have the slope, you can use the point-slope formula, y - y1 = m(x - x1), where (x1, y1) is the given point, to find the equation of the tangent line.

4. Can I use the tangent line to approximate the value of a function?

Yes, the tangent line can be used to approximate the value of a function at a specific point. This is known as linear approximation and involves using the tangent line to estimate the value of the function at a given x-value. The closer the given x-value is to the point of tangency, the more accurate the approximation will be.

5. What is the relationship between slope and the tangent line?

The slope of a line is equal to the slope of the tangent line at any point on a curve. This is because the tangent line is a straight line that touches the curve at only one point, so its slope is equal to the slope of the curve at that point. This relationship is important in determining the behavior of a curve and in finding the equation of the tangent line.

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