Determining values of cubic function with conditions (calculus)

Question

Question: Consider the cubic function $f(x) = ax^3 + bx^2 + cx + d$. Determine the values of $a$, $b$, $c$, and $d$ so that all of the following conditions are met.

a) $′(−1) = 1, ′(0) = −2$,

b) There is a point of inflection at $(1, 0)$

c) The function $y = f(x)$ intercepts y axis at $(0, 1)$

Find the constants $a$, $b$, $c$, and $d$.

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This question is really confusing to me as I am still unaware on what the point of inflection is and how to calculate it. Along with that, I don't know how to solve this question. Which formula should I use and what steps do I need to meet the following conditions. Please help. This question is extremely confusing. Thanks!

Answer 1

find f'$$f(x) = ax^3 + bx^2 + cx + d\\\to f'=3ax^2+2bx+c\\\to f''=6ax+2b$$now you have 4 variable and 4 information @ the question
$$f'(-1)=1 \to 3ax^2+2bx+c|_{x=-1}=1\\f'(0)=-2 \to 3ax^2+2bx+c|_{x=0}=-2$$ for inflection point you will have $$f''(x)=0\to 6ax+2b|_{x=1}=0$$ and y-intercept $$f(x) = ax^3 + bx^2 + cx + d|_{x=1}=0$$ these all mean $$\left\{ \begin{array}{l} 3a-2b+c=1\\ 0(a)+0(b)+c=-2\\\ 6a(1)+2b=0\\ a+b+c+d=0 \end{array} \right.$$

Answer 2

If there is a point of inflexion at $x=1$ then $$f''(x)=k(x-1) \implies f'(x)=kx^2/2-kx+p \implies f(x)=kx^3/6-kx^2/2+px+q$$ Next, $f'(-1)=1$ gives $1=k/2+k+p=1$, $f'(0)=-2$ gives $p=-2$, we get $k=2$. Lastly, we use $f(0)=1$, we get $q=1$. So finally t6he cubic is $$f(x)=\frac{x^3}{3}-x^2-2x+1.$$

Answer 3

We can also approach this sort of question by using the information provided to solve for coefficients as we go along, so as to avoid having a large system of equations to solve.

We are told outright that the curve for the polynomial has its $ \ y-$intercept at $ \ (0 \ , \ 1) \ $ , so we have $ \ a·0^3 + b·0^2 + c·0 + d \ = \ \mathbf{d \ = \ 1} \ \ . $

The information about one of the points of inflection being at $ \ ( 1 \ , \ 0 ) \ $ also gives us one of the $ \ x-$ intercepts of the curve. As a point of inflection, we have that $$ \ f''(1) \ \ = \ \ 6a·1 \ + \ 2b \ \ = \ \ 0 \ \ \Rightarrow \ \ b \ \ = \ \ -3a \ \ . $$

The two values of the slope of the tangent lines to the curves then tell us that $$ f'(-1) \ \ = \ \ 3a·(-1)^2 \ + \ 2b·(-1) + c \ \ = \ \ 3a·1 \ + \ 2·(-3a)·(-1) \ + \ c \ \ = \ \ 9a \ + \ c \ \ = \ \ 1 \ \ ; $$ $$ f'(0) \ \ = \ \ 3a·0^2 \ + \ 2b·0 \ + \ c \ \ = \ \ \mathbf{c \ \ = \ \ -2} \ \ . $$

We can now solve the first of this pair of equations for $ \ a \ \ : \ \ 9a + c \ = \ 9a + (-2) \ = \ 1 $

$ \Rightarrow \ \ 9a \ = \ 3 \ \ \Rightarrow \ \ \mathbf{a \ = \ \frac13} \ \ , $ which leads immediately to $ \ b \ = \ -3a \ \Rightarrow \ \mathbf{b \ = \ -1} \ \ . $ The cubic polynomial is thus $ \ f(x) \ = \ \mathbf{\frac13 x^3 \ - \ x^2 \ - \ 2x \ + \ 1} \ \ . $ This agrees with Z Ahmed's result.

However, when we check this against the statement concerning the $ \ x-$ intercept at $ \ (1 \ , \ 0) \ \ , $ we find

$$ f(1) \ = \ \frac13 · 1^3 \ - \ 1^2 \ - \ 2·1 \ + \ 1 \ \ \neq \ 0 \ \ . $$

So there is an inconsistency in the specification of the conditions for this problem; a graph of the polynomial will confirm this. My suspicion is that the poser changed one of the details along the way and neglected to check whether anything else was affected, or made a miscalculation in preparing the problem. (Not that I've ever done something like that in making up homework sets or an exam... hrmm...)