Finding f'(1) with Tangent Line at (1,7)

[lude1]

Homework Statement

If the line tangent to the graph of the function f at the point (1,7) passes through the point
(-2,-2), then f'(1) is?

Answer: 3

Homework Equations

The Attempt at a Solution

I got the right answer, but I think it was by pure luck. I found the slope using this formula:

y-y₁ = m(x-x₁)​

Then I plugged everything in, getting:

7 + 2 = m(1 + 2)
9 = m3
m = 3​

I originally created a new equation with this slope and (-2,-2) and did the derivative, but that's just rhetorical since the slope is 3 anyway. I guess I'm having a problem finding the equation of the derivative? Since once I have that equation, I can just plug in one. Or was my approach correct and the answer was y' = 3 (therefore no matter what point I plug in, the answer is 3 anyway)

 

[Mark44]

Homework Statement

If the line tangent to the graph of the function f at the point (1,7) passes through the point
(-2,-2), then f'(1) is?

Answer: 3

Homework Equations

The Attempt at a Solution

I got the right answer, but I think it was by pure luck. I found the slope using this formula:

y-y₁ = m(x-x₁)​

No, this is pretty much what you should have done. You have a line through (1, 7) and (-2, -2), so the slope of the line is 3, and this is f'(1).

Then I plugged everything in, getting:

7 + 2 = m(1 + 2)
9 = m3
m = 3​

I originally created a new equation with this slope and (-2,-2) and did the derivative, but that's just rhetorical since the slope is 3 anyway. I guess I'm having a problem finding the equation of the derivative?

You can't find the general equation of f'(x), since you don't know the equation of f(x). All you know about f is that its derivative at x = 1 is 3.

Since once I have that equation, I can just plug in one. Or was my approach correct and the answer was y' = 3 (therefore no matter what point I plug in, the answer is 3 anyway)

y' = 3 is not correct. You can say that y'(1) = 3 or f'(1) = 3, but you can't say anything about the derivative of this function at an arbitrary point.

 

[lude1]

So "the line tangent to the graph of function f" means the derivative. "At the point (1,7) and passes through the point (-2,-2)" means the derivative passes through these two points.

Therefore, to find the derivative or the slope, you use the equation y-y₁ = m(x-x₁).

The reason why we know f'(1) = 3 is because "the line tangent to the graph of the function f at the point (1,7)" or the derivative to the graph of f at points (1,7) and (-2,-2) is 3, correct?

Sorry! I'm having a hard time wrapping my head around this despite how easy it is

 

[Mark44]

So "the line tangent to the graph of function f" means the derivative.

Not exactly. The slope of the line tangent to the graph of f is the derivative. A better way to say this is that at the point (x, f(x)), the slope of the tangent line is given by f'(x).

"At the point (1,7) and passes through the point (-2,-2)" means the derivative passes through these two points.

No, the derivative is a function. The tangent line at the point (1, 7) passes through (-2, -2).

Therefore, to find the derivative or the slope, you use the equation y-y₁ = m(x-x₁).

Better yet is the formula y₂-y₁ = m(x₂-x₁), or m = (y₂-y₁)/(x₂-x₁).

The reason why we know f'(1) = 3 is because "the line tangent to the graph of the function f at the point (1,7)" or the derivative to the graph of f at points (1,7) and (-2,-2) is 3, correct?

Almost. Here's what you know.
1. The line through (1, 7) and (-2, -2) is tangent to the graph of f at (1, 7).
2. The slope of this line is 3.
3. From 1 and 2, the slope of the tangent line at (1, 7) is 3.
4. From 3, f'(1) = 3.

Sorry! I'm having a hard time wrapping my head around this despite how easy it is

 

[lude1]

Ooh, okay. I get it now! Thank you so much!