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Question 1190587: Find the tangent to f(x) = x^2 at the point where
i. x= 1
ii. x= 4
Please explain step-by-step!
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part i)
Use the power rule to get the derivative
f(x) = x^2
f ' (x) = 2x
The derivative function will directly determine the slope of the tangent line.
If x = 1, then
f ' (x) = 2x
f ' (1) = 2*1
f ' (1) = 2
The slope of the tangent is m = 2 when x = 1.
Plug x = 1 into the original function
f(x) = x^2
f(1) = 1^2
f(1) = 1
The point (1,1) is on the f(x) curve.
Now apply the point-slope form
y - y1 = m(x - x1)
y - 1 = 2(x - 1)
y - 1 = 2x - 2
y = 2x-2+1
y = 2x-1
The slope of the tangent line is y = 2x-1 when x = 1
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For part ii), you'll follow very similar steps.
Use f ' (x) = 2x from earlier.
You should get a tangent slope of m = 8 and the point (4,16) is on the parabola.
The equation of the tangent line is y = 8x-16 when x = 4.
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