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

Answer:

y = 1.

Step-by-step explanation:

f(x) = x^2 + 1

We know that:

$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

replacing the function, we get:

$f'(x) = \lim_{h \to 0} \frac{(x + h)^2 + 1 - (x^2 + 1)}{h} \\ = \lim_{h \to 0} \frac{(x + h)^2 - x^2}{h} \\\\ = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h}$

Now we can simplify this to:

$= lim_{h \to 0} \frac{ 2xh + h^2 }{h}$

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.