Question

If f is a second-degree polynomial function such that f(3)=5, f'(3)=6, and f''(3)=4, what is the value of f(2) ?

Answer 1

so let's say the equation is y = ax² + bx + c, where a,b,c are constants.

we know that f(3) = 5, or namely when x = 3 , y = 5.

we also know that f'(3) = 6, or namely the slope at the point (3,5) is 6.

we also know that at (3,5) the 2nd derivative is 4, so a positive number simply tell us its concavity, is up, so is a parabola opening upwards.

`y=ax^2+bx+c\implies \left. \cfrac{dy}{dx}=2ax+b \right|_(x=3)~~ = ~~ 6\implies 6=2ax+b \\\\\\ 6=2a(\stackrel{x}{3})+b\implies 6=6a+b\implies 6-6a=b\implies \boxed{6(1-a)=b} \\\\[-0.35em] ~\dotfill\\\\ \left. \cfrac{d^2y}{dx^2}=2a \right|_(x=3)~~ = ~~4\implies 2a=4\implies a=\cfrac{4}{2}\implies \boxed{a=2}~\hfill \boxed{-6=b} \\\\[-0.35em] ~\dotfill`

`y=2x^2-6x+c\qquad \begin{cases} x=3\\ y = 5 \end{cases}\implies 5=2(3)^2-6(3)+c\implies \boxed{5=c} \\\\\\ y=2x^2-6x+5~\hfill y(2)=2(2)^2-6(2)+5\implies \blacktriangleright y(2)=1 \blacktriangleleft`