Answer:
y = 1.
Step-by-step explanation:
f(x) = x^2 + 1
We know that:
[tex]f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}[/tex]
replacing the function, we get:
[tex]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} \\\[/tex]
Now we can simplify this to:
[tex]= lim_{h \to 0} \frac{ 2xh + h^2 }{h}[/tex]
and solving the quotient, we get:
[tex]= lim_{h \to 0} 2x + h = 2x + 0 = 2x[/tex]
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.
