of the tangent to the curve at the point (4, 3/4)
and the equation of the normal to the curve at this point
You don't say what level math you are in, but this is most easily solved using calculus so I hope you have reached that level.
the first derivative is defined as the slope (gradient) of the tangent line to a curve.
y = 1/√x + 1/x
write with exponents to take the derivative using the power rule:
y = x^(-1/2) + x^-1
derivative:
y ' = (-1/2)x^(-3/2) - x^-2
now plug in the x-coordinate for the given point to find the slope:
m = (-1/2)(1/√4^3) - 1/4^2
= -1/16 - 1/16 = -1/8
use point-slope form to get equation:
y - 3/4 = (-1/8)(x - 4)
simplify to preferred form.
to find the equation of the normal (perpendicular) line, just use the negative reciprocal of the slope (8) and follow the same steps.
that's it! ;)
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You don't say what level math you are in, but this is most easily solved using calculus so I hope you have reached that level.
the first derivative is defined as the slope (gradient) of the tangent line to a curve.
y = 1/√x + 1/x
write with exponents to take the derivative using the power rule:
y = x^(-1/2) + x^-1
derivative:
y ' = (-1/2)x^(-3/2) - x^-2
now plug in the x-coordinate for the given point to find the slope:
m = (-1/2)(1/√4^3) - 1/4^2
= -1/16 - 1/16 = -1/8
use point-slope form to get equation:
y - 3/4 = (-1/8)(x - 4)
simplify to preferred form.
to find the equation of the normal (perpendicular) line, just use the negative reciprocal of the slope (8) and follow the same steps.
that's it! ;)