MHB Katie's Question: Find f(2) & f'(2) of Tangent Line y=4x-5

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To find f(2) and f'(2) for the tangent line y=4x-5, it is established that the slope of the tangent line, which is 4, must equal the derivative of the function f(x) at x=2. Therefore, f'(2) is determined to be 4. Additionally, to find f(2), the value of the tangent line at x=2 is calculated as y(2)=4(2)-5, resulting in f(2)=3. The function f(x) is confirmed to be f(x)=x^2-1, which aligns with the tangent line's properties at the point of tangency. The solution effectively demonstrates the relationship between the function and its tangent line at the specified point.
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Here is the question:

If an equation of the tangent line to the curve y=f(x) at the point where a=2 is...?


y=4x-5, find f(2) and f'(2)

Please walk me through this question step by step.

I have posted a link there to this topic so the OP can see my work.
 
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Hello Katie,

Consider the following diagram, showing the function $f(x)=x^2-1$ and its tangent line $y=4x-5$:

View attachment 1403

As you can see, the instantaneous slope of the function must match the slope of the tangent line at the point of tangency, and the value of the function at that point must be equal to the value of the tangent line at that point since they touch there. Hence, we must have:

$$f'(2)=\frac{d}{dx}(4x-5)=4$$

$$f(2)=y(2)=4(2)-5=3$$
 

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