y'=x∛y
The tangent line at (2,8) is thus
y-8 = 4(x-2)
y'=x∛y
y^(-1/3) dy = x dx
3/2 y^(2/3) = 1/2 x^2 + c
y^(2/3) = 1/3 x^2 + c
since y(2) = 8,
4 = 4/3 + c
c = 8/3
y^(2/3) = (x^2 + 8)/3
The domain is clearly where x^2+8 >= 0, or all reals.
for minimum value, y'=0, or where x=0 or y=0. Since y is never zero, the minimum is at x=0. Just plug it in.
The expression dy/dx = x*(cube root y) gives the slope at any point of the graph of the function f(x) where f(2) = 8.
A. Write the equation of the tangent line to f(x) and any point (2,8).
B. Write an expression for f(x) in terms of x.
C. What is the domain of f(x)?
D. What is the minimum value of f(x)?
Thank you so much!!
1 answer
1 answer