Thread: Derivatives about the Tangent Line

Find an equation of the tangent line to the the graph of the function
f(x)=(x^(1/2))(3-8x^2)/x

when x=1.

Answer: We note that, when x=1, the corresponding point on the graph is given by (1,f(1))=(1,_____), and that the slope of the tangent line
is f ¢(1)= _____.

Therefore the equation of the tangent line is y =______

Originally Posted by shannon1111

Find an equation of the tangent line to the the graph of the function
f(x)=(x^(1/2))(3-8x^2)/x

when x=1.

Answer: We note that, when x=1, the corresponding point on the graph is given by (1,f(1))=(1,_____), and that the slope of the tangent line
is f ¢(1)= _____.

Therefore the equation of the tangent line is y =______

The first thing I would do is rewrite the function as f(x)(x^(1/2))(3-8x^2)x^(-1)= x^(-1/2)(3-8x^2)= 3x^(-1/2)- 8x^(3/2).

Surely, you can do the first part: when x= 1, what is f(1)? That's just arithmetic.

To find the slope, take the derivative of f. Use the fact that the derivative of Ax^n+ Bx^m is nAx^(n-1)+ mBx^(m-1).