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QuestionFind the tangent line to f(x)=−x2+5f(x)=-x^{2}+5 at x=5x=5 and use it to estimate f(5.1)f(5.1).

Studdy Solution

STEP 1

Assumptions1. The function is f(x)=−x+5f(x)=-x^{}+5 . The point of tangency is x=a=5x=a=5
3. The tangent line uses the instantaneous rate of change (the derivative of the function) at the point x=ax=a for the slope

STEP 2

First, we need to find the derivative of the function f(x)f(x), which represents the instantaneous rate of change of the function. The derivative of f(x)f(x) is denoted as f′(x)f'(x).
f′(x)=ddx(−x2+5)f'(x) = \frac{d}{dx}(-x^{2}+5)

STEP 3

Now, calculate the derivative of f(x)f(x).
f′(x)=−2xf'(x) = -2x

STEP 4

Now that we have the derivative, we can find the slope of the tangent line at x=a=x=a= by substituting x=x= into f′(x)f'(x).
f′()=−2()f'() = -2()

STEP 5

Calculate the slope of the tangent line.
f′(5)=−2(5)=−10f'(5) = -2(5) = -10

STEP 6

We also need the y-coordinate of the point of tangency. We can find this by substituting x=5x=5 into f(x)f(x).
f(5)=−(5)2+5f(5) = -(5)^{2}+5

STEP 7

Calculate the y-coordinate of the point of tangency.
f(5)=−(5)2+5=−20f(5) = -(5)^{2}+5 = -20

STEP 8

Now that we have the slope of the tangent line and the point of tangency (5,−20)(5, -20), we can write the equation of the tangent line in point-slope form y−y1=m(x−x1)y - y_{1} = m(x - x_{1}), where mm is the slope and (x1,y1)(x_{1}, y_{1}) is the point of tangency.
y−(−20)=−10(x−5)y - (-20) = -10(x -5)

STEP 9

implify the equation of the tangent line.
y+20=−x+50y +20 = -x +50

STEP 10

Rearrange the equation to get it in slope-intercept form y=mx+by = mx + b.
y=−10x+30y = -10x +30

STEP 11

Now that we have the equation of the tangent line, we can use it to approximate f(5.)f(5.) by substituting x=5.x=5. into the equation of the tangent line.
y=−10(5.)+30y = -10(5.) +30

STEP 12

Calculate the approximation for f(5.)f(5.).
y=−10(5.)+30=−21y = -10(5.) +30 = -21So, the tangent line approximates that f(5.)f(5.) is −21-21.

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