Determine the point at which the graph of the function has a horizontal tangent line

Question

Answer 1

Find f'(x), set that that to zero and solve for x

f'(x)=(10x(x²+5)-10x³)/(x²+5)² = 50x/(x²+5)²

50x/(x²+5)² = 0

x=0

f(0) = 0

The graph has a horizontal tangent line at the point (0,0).

Answer 2

take the derivative and set its numerator = zero

derivative of a u/v = (vu'-uv')/v^2 u=5x^2, v= x^2 +5, u'= 10x, v'= 2x

f'(x) = (x^2+5)10x - 5x^2(2x) all over [x^2+5]^2

set the numerator = 0

10x^3+50x -10x^3 = 0

50x =0

x=0

point of tangency with a horizontal line is the origin (0,0)

another way to calculate the derivative

is f(x) = 5x^2(x^2+5)^-1

f(x) = uv, f'(x) = uv' + vu' u=5x^2, u'=10x, v= (x^2+5)^-1, v'= -2x(x^2+5)^-2

f''(x) = 5x^2(-2x)(x^2+5)^-2 + (x^2+5)^-1(10x) = -10x^3/(x^2+5) + 10x^3+50x/(x^2+5)^2

= 50x/(x^2+5)^2