Let f(x)=ax^3+bx^2+cx+41 be such that f(1)=40,f^'(1)=2 and f^''(1)=4.

Question

Let $f(x)=a{x}^{3}+b{x}^{2}+cx+41$ be such that $f(1)=40,{f}^{'}(1)=2$ and ${f}^{''}(1)=4$. Then ${a}^{2}+{b}^{2}+{c}^{2}$ is equal to:

Accepted Answer

Given the polynomial function:

$f(x) = ax^3 + bx^2 + cx + 41$

We are provided the following conditions from the problem:

1. $f(1) = 40$

2. $$f^{\prime}(1) = 2$$

3. $$f^{\prime \prime}(1) = 4$$

First, calculate $f(1)$:

$f(1) = a(1)^3 + b(1)^2 + c(1) + 41 = 40$

Simplifying, we get:

$a + b + c + 41 = 40$

Therefore:

$a + b + c = -1$

Next, calculate the first derivative $f^{\prime}(x)$:

$$f^{\prime}(x) = 3ax^2 + 2bx + c$$

Given $f^{\prime}(1) = 2$:

$$f^{\prime}(1) = 3a(1)^2 + 2b(1) + c = 2$$

Simplifying, we get:

$3a + 2b + c = 2$

Next, calculate the second derivative $f^{\prime \prime}(x)$:

$$f^{\prime \prime}(x) = 6ax + 2b$$

Given $f^{\prime \prime}(1) = 4$:

$$f^{\prime \prime}(1) = 6a(1) + 2b = 4$$

Simplifying, we get:

$6a + 2b = 4$

Dividing the entire equation by 2:

$3a + b = 2$

We now have three equations:

1. $a + b + c = -1$

2. $3a + 2b + c = 2$

3. $3a + b = 2$

To solve for $a$, $b$, and $c$, follow these steps:

First, subtract the third equation from the second equation:

$(3a + 2b + c) - (3a + b) = 2 - 2$

Which simplifies to:

$b + c = 0$

So,

$c = -b$

Substitute $c = -b$ into the first equation:

$(a + b - b = -1)$

Simplifying, we get:

$a = -1$

Now substitute $a = -1$ into the third equation:

$3(-1) + b = 2$

Which simplifies to:

$-3 + b = 2$

Therefore:

$b = 5$

Next, since $c = -b$:

$c = -5$

Finally, we need to find $a^2 + b^2 + c^2$:

$a^2 + b^2 + c^2 = (-1)^2 + 5^2 + (-5)^2$

Simplifying, we get:

$1 + 25 + 25 = 51$

Therefore, the answer is:

Option B: 51

Solutions · 1