Equation of a Tangent line

Question

Hi can someone walk me through all of the steps of solving this? I'm completely stuck

Answer 1

Sure thing.

So we have our equation f(x) = (4 - x³)/(x+ x²)

How do we find the equation of a tangent line? Well, we need the slope and a point.

The point part is easy. We are given x=1 as our point of tangency, so we can plug in 1 for x to find the y coordinate. It is simply 3/2 = 1.5. So, our point is (1,1.5).

Now, to find the slope, we need to take the derivative of our function and plug in x = 1. This is because derivatives are rates of change, which is a slope.

So, we need to find f'(x). For this, use quotient rule.

f'(x) = [(x+x²)(-3x²)-(4-x³)(1+2x)]/(x²+x)²

f'(x) = -[x⁴+2x³+8x+4]/x²(x+1)² after you have simplified.

f'(1) = -15/4

So, we now have a point and a slope. Our slope is -15/4 and our point is (1,1.5). We can use point-slope form to find out our final line now.

y-1.5 = -15/4(x-1)

Let me know if you have any other questions!