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Can you solve this question?
y=?
(?,?)

Can you solve this question? y=? (?,?)-example-1

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Answer:

The tangent line is: y = 8x - 58

The tangent point is at (7, -2)

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Step-by-step explanation:

One of the definitions of derivatives is


\displaystyle f'(a) = \lim_(x\to a) (f(x)-f(a))/(x-a)

where f ' (a) represents the derivative evaluated at x = a.

The value of f ' (a) will get us the slope of the tangent at x = a.

The idea is that x is getting closer and closer to 'a'. In doing so, the secant lines slowly approach the tangent line.

Keep in mind that x will never reach 'a' itself (if it did, then we'd have a division by zero error).

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The given limit we have is


\displaystyle \lim_(x\to 7) (f(x)+2)/(x-7) = 8

and that is equivalent to


\displaystyle \lim_(x\to 7) (f(x)-(-2))/(x-7) = 8

and also equivalent to


\displaystyle \lim_(x\to 7) (f(x)-f(7))/(x-7) = 8

Compare that to the template I mentioned at the top to see that

  • a = 7
  • f(a) = f(7) = -2
  • f ' (a) = f ' (7) = 8

Therefore, we can say the tangent slope is 8 and the tangent touches the f(x) curve at (x,y) = (a, f(a)) = (7,f(7)) = (7,-2)

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m = 8 = \text{slope}\\\\(x_1,y_1) = (7,-2) = \text{tangent point}\\\\

Let's use that info to determine the equation of the tangent line.

I'll use point-slope form to isolate y.


y-y_1 = m(x-x_1)\\\\y-(-2) = 8(x-7)\\\\y+2 = 8x-56\\\\y = 8x-56-2\\\\y = 8x-58\\\\

That's the equation of the tangent line to the point (7,-2).

User Ilalex
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