Application of Derivatives : Tangent to a parabola

Question

I am trying to solve the question in the image above. Here is how I tried :

I am stuck at this point.How should I solve I further? AND Second derivative is independent of x then how to decide between f"(0) and f"(1).

Answer 1

I'll do it your way first.

Following what you wrote, you know that $f'(x) = \frac{x-h}{2a}$, and you know that $f''(x) = \frac{1}{2a}.$ Also, near the bottom of the first page you have written that $$1-h=2a.\tag1$$ This is important, because it is equivalent to the equation $\frac{1-h}{2a}=1,$ which is equivalent to the fact that $f'(1) = 1,$ which is what we learn from the fact that the parabola is tangent to the line $y=x$ at $x=1.$

At the very bottom of the page you have found that $$f'(1) + f''(x) = \frac{2-h}{2a}.\tag2$$ (I have translated what you wrote from the Leibniz notation that you used into the notation used in the question.)

Notice that there is only one choice among the multiple-choice answers that matches Equation $2,$ and that's choice (d). But if choice (d) is the correct answer, then $f'(1) + f''(x) = f''(0) + f'(1) = 1,$ which implies that $\frac{2-h}{2a} = 1,$ which implies $2-h=2a.$ But you already know that $1-h=2a$ (Equation $1$). Is it possible both equations are true? Is it possible that $2-h=1-h$? No, so (d) is not the correct answer.

If you write out an equation for choice (b), you end up with $\frac{1}{2a} - \frac{1-h}{2a} = 1.$ Again, you also know that $\frac{1-h}{2a} = 1.$ Can this be true? Yes, if $h = \frac12.$ But how can you show that $h=\frac12$? Hmm, not looking good, maybe we'll come back to this later, but let's give the other choices a chance.

Choices (a) and (c) evaluate the derivatives at different points than (b) and (d): $f''(1)$ and $f'(0).$ But you already know that $f''(1)= \frac{1}{2a}$ and $f'(0) = \frac{0 - h}{2a} = -\frac{h}{2a}.$ Try this with choice (a), which says that $f''(1) + f'(0) = 1.$ Is it true that $\frac{1}{2a}+\left(-\frac{h}{2a}\right)=1$? Yes, if $\frac{1-h}{2a}=1.$ But we know this is true, because Equation $1$ is true. So (a) is correct.

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I consider that the hard way to do this problem. The easy way is, since I know $f''(x)$ is constant, I know $y=f'(x)$ is the equation of a line of slope $m = f''(x).$ Writing the equation of this line in $y$-intercept form, I find that
$$f'(x) = f'(0) + m x.\tag3$$ I happen to notice that $f'(1) = 1,$ since the parabola is tangent to a line of slope $1$ at $x=1,$ so putting $x=1$ in Equation $3$ I have $$1 = f'(0) + m,$$ where $m$ is the second derivative of $f$ evaluated at any point. I could replace $m$ with $f''(0)$ or with $f''(1),$ whichever is more convenient. Looking at the choices, I think $f''(1)$ is more convenient, because then I have $1 = f'(0) + f''(1),$ so I know that at least choice (a) is correct.

Answer 2

$y(1) = 1\\ y = (x-1)(ax+b) + 1$

$y'(x) = (ax+b) + a(x-1)\\ y'(1) = 1\\ b = 1-a\\ y'(0) = 1-2a$

$y''(1) = y''(0) = 2a$

a) is the only one that fits.

Answer 3

General form of parabola $$f(x) = ax^2 + bx + c$$

We are given that the line $y=x$ touches the parabola at $x=1$, which implies that the gradient of the parabola at $x=1$ is given by the gradient of $y=x$

Hence $f'(1) = 1$, and $f(1) = 1$

Plugging this in, we get:

$$f'(x) = 2ax + b \implies f'(1) = 2a + b = 1 \implies b=1-2a$$

$$f(1) = a+b+c = 1$$

Also, $$f''(x) = 2a$$

So then $$f''(x)-f'(x) = 2a-2ax-b = 2a-2ax+2a-1 =4a-2ax-1$$

$$f'(x)+f''(x) = 2ax+b+2a = 2ax+1-2a+2a = 2ax+1$$

It seems like $f'(x)+f''(x)$ will fit one of those expressions.

Also note that $f''(x)$ is independent of $x$

$$f''(1)+f'(0) = 1$$

Note that $$f''(0)+f'(1) =2a+1$$ which doesn't necessarily equal $1$

Therefore the answer is $(a)$