## University Calculus: Early Transcendentals (3rd Edition)

At 2 points $(2,4)$ and $(4,16)$, the tangent lines to $f(x)$ pass through the point $(3,8)$.
$$f(x)=x^2$$ 1) Find $f'(x)$: $$f'(x)=2x$$ 2) The slope of the tangent line to $f(x)$ at $x=a$ is $f'(a)=2a$. We take the point $A(a, a^2)$ to be the point of tangency of a tangent line $(t)$ to $f(x)$. Now, take $(t)$ to be passing through $B(3,8)$. With these 2 points $A$ and $B$, we can have a formula of the slope $s_t$ of $(t)$: $$s_t=\frac{a^2-8}{a-3}$$ But $s_t=f(a)=2a$ as well. That means, $$\frac{a^2-8}{a-3}=2a$$ $$a^2-8=2a^2-6a$$ $$a^2-6a+8=0$$ $$(a-2)(a-4)=0$$ $$a=2\hspace{1cm}\text{or}\hspace{1cm}a=4$$ - For $a=2$: $$f(a)=2^2=4$$ - For $a=4$: $$f(a)=4^2=16$$ Therefore, at 2 points $(2,4)$ and $(4,16)$, the tangent lines to $f(x)$ pass through the point $(3,8)$.