What is the equation of a line tangent to a curve at this given point:

Question

f(x)= ((sqrt x)+3)(x^2-5) at x=1

Needing to know how to find this equation. Any help would be greatly appreciated. Please also show your work. Thanks!

Answer 1

Step 1: Find the derivative function of f(x) using the product rule: (f(x)g(x))'=f'(x)g(x)+f(x)g'(x):

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

Step 2: Evaluate f'(x) at x=1 to get the slope f(1)=(1/2)(1-5)+(1+3)(2)=-2+8=6

Step 3: Find the y=f(1)=(1+3)(1-5)=4(-4)=-16, thus the point is (1,-16)

Step 4: Use the slope-point formula to obtain the tangent line at x=1:

y-(-16)=6(x-1)=6x-6

y+16=6x-6

Thus the tangent line at x=1 is y=6x-22 or y-6x+22=0.