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Calculus help please?

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5x^3+radicalx (1,6) How do I write an equation for this? (tangent line)

asked Nov 12, 2014 in PRECALCULUS by anonymous

1 Answer

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The function is f(x) = 5x³ + √x .

Slope of the tangent line is f '(x) at x = 1 .

f '(x) = 5*3x² + 1/2√x                     [ derivative of  xn = n xn-1 and deriavtive of √x is  1/2√x  ]        

f '(x) = 15x² + 1/2√x  

f '(1) = 15(1)² + 1/2√1 

f '(1) = 15 + 1/2

f '(1) = 31/2 

So the slope of tangent is 31/2 .

The equation of tangent with slope 31/2  and point (1,6) using slope point form .

y – y1 = m(x – x1)

y – 6 = 31/2 (x – 1)

2(y - 6) = 31(x – 1)

2y -12 = 31x -31

31x -2y -19 = 0

So the equation of tangent is 31x -2y -19 = 0 .

answered Nov 12, 2014 by yamin_math Mentor

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