Coincide Tangent Lines

Let the common tangent be $y=mx+n$ . As it intersects with $y=f(x)$ , we consider the equation $f(x)=mx+n$ , or equivalently $x^4+ax^3+bx^2+(c-m)x+(d-n)=0$

From the conditions, we know that $x=1$ and $x=5$ are the repeated roots, so the above equation can be written as $\left(x-1\right)^2\left(x-5\right)^2=0$ . Expand it and we will obtain $x^4-12x^3+46x^2-60x+25=0$ .

Compare the coefficients of $x^3$ and $x^2$ , we have $a=-12$ and $b=46$ . Then $a+b=\boxed{34}$ .

Note that the values of $c$ and $d$ may be arbitrary.

Let the gradient of the straight line and hence that at $(1, f(1))$ and $(5, f(5))$ be $m$ . Then

$\begin{cases} m = \dfrac {f(5)-f(1)}{5-1} = 156 + 31a + 6b + c & ...(1) \\ m = f'(1) = 4 + 3a + 2b + c & ...(2) \\ m = f'(5) = 500+ 75a + 10b + c & ...(3) \end{cases}$

$\implies \begin{cases} (1) = (2): \quad 7a + b = - 38 & ...(4) \\ (2)=(3): \quad 9a + b = - 62 & ...(5) \end{cases}$

From $(5) - (4): \ 2a = -24 \implies a = - 12$ . From $(2): \ - 84 + b = -38 \implies b = 46$ . Therefore, $a+b = -12+46 = \boxed{34}$ .

I calculated the first derivative and set them to be equal to each other

$f'(x) = 4x^3 + 3ax^2 + 2bx + c$

$f'(1) = 4 + 3a + 2b + c$

$f'(5) = 500 + 75a + 10b + c$

$500 + 75a + 10b + c = 4 + 3a + 2b + c$

I then calculated the second derivative and set them to be equal to each other.

$f''(x) = 12x^2 + 6ax + 2b$

$f''(1) = 12 + 6a + 2b$

$f''(5) = 300 + 30a + 2b$

From here, we solve the system of equations
$500+75a+10b+c = 4+3a+2b+c$

$300 + 30a + 2b = 12 + 6a + 2b$

We get
$a=-12$

$b = 46$

$a + b = \boxed{34}.$

Would this method always work?

$f(x) = x^4 + ax^3 + bx^2 + cx + d \implies f'(x) = 4x^3 + 3ax^2 + 2bx + c$

$f'(5) = 500 + 75a + 10b + c$ and $f'(1) = 4 + 3a + 2b + c$ and using the coordinates of the given points the slope

$m = \dfrac{624 + 124a + 24b + 4c}{4}$

$f'(5) = m = f'(1) \implies 2000 + 300a + 40b = 624 + 124a + 24b$ and $16 + 12a + 8b = 624 + 124a + 24b$

$\implies$

$11a + b = -86$

$-7a - b = 38$

$\implies a = -12$ and $b = 46 \implies a + b = \boxed{34}$ .

We know that f'(1) = f'(5). Solving this equation you obtain that 9a+b = -62. We know from the graph attached to the task that f(1) and f(5) lay on the same line. Thanks to it we can calculate f(5) as f(1)+4 f'(1). If you compare f(5) with f(1)+4 f'(1) you will obtain another relationship between a and b - 7a+b = -38. Now you have two equations from which you can easily calculate that a = -12 and b = 46, so a+b = 34.

If we can find ONE function with this property, we can get a+b from that. The simplest is

$f(1)=f(5)=0$ and

$f'(1)=f'(5)=0$

If a polynomial has a factor (x-n)^2, then its value and slope are 0 at x=n. Thus,

$f(x) = (x-1)^2 * (x-5)^2$

works here. It expands to

$f(x)=x^4 - 12x^3 + 46x^2 - 60x + 25$

where a+b = 34

We can then generalize to say that ANY polynomial with a=-12, b=46 has this property, and no others do.

• $f(x) + d$ would just translate f(x) upwards, preserving the tangent line. Thus d can have any value

• $f(x) + cx$ would change the tangent line's slope, but preserve the line. Thus c can have any value

• $f(x) + b'x^2, f(x) + a'x^3$ (where a' and b' are different from a and b) would change the function's slope unevenly at every point, so that f'(1) would not equal f'(5):

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

$f'(5) = 25a' + 5b'$

There is no combination of a' and b' that would make these values of f'(x) equal except a'=b'=0. a and b MUST be exactly -12 and 46 respectively.

It's easy to see that the dotted line corresponds to $g(x) = x$ and the problem is telling us that $f$ and $g$ are tangent at $x=1, 5$ (i.e. each of these values should be a double root of the difference of the polynomials). So we have that $f-g$ is a monic polynomial (leading coefficient is 1) that has a double root at 1 and at 5. Therefore:

$f(x) - g(x) = 1\times(x-1)^2(x-5)^2 = x^4-12x^3+46x^2-59x+25$

And since $g(x) = x$ we see that $a = -12, b = 46$ . Therefore $a + b = 34$