Question

Find an equation of the tangent line to the curve,f(x) =1x2+ 1at (0,1) by using the limitdefinition of the derivative. (Dont use Derivative Rules)

Answer 1

y = 1.

Explanation:

f(x) = x^2 + 1

We know that:

`f'(x) = \lim_(h \to 0) (f(x + h) - f(x))/(h)`

replacing the function, we get:

`f'(x) = \lim_(h \to 0) ((x + h)^2 + 1 - (x^2 + 1))/(h) \\ = \lim_(h \to 0) ((x + h)^2 - x^2)/(h) \\\\ = \lim_(h \to 0) (x^2 + 2xh + h^2 - x^2)/(h) \\\`

Now we can simplify this to:

`= lim_(h \to 0) ( 2xh + h^2 )/(h)`

and solving the quotient, we get:

`= lim_(h \to 0) 2x + h = 2x + 0 = 2x`

then f'(x) = 2*x

And we want the equation for the tangent line at (0, 1)

The slope of that line will be:

f'(0) = 2*0 = 0

So this is a horizontal line, of the type y = a

And we know that this line must pass through the point (0, 1)

so we must have y = 1.

Then the equation of the tangent line to f(x) at the point (0, 1) is:

y = 1.