Let $f(x)=ax^3+bx^2+cx+41$ be such that $f(1)=40$, $f'(1)=2$ and $f''(1)=4$. Then $a^2+b^2+c^2$ is equal to:
Step-by-Step Solution
Key Concept: $f'(x)=3ax^2+2bx+c$, $f''(x)=6ax+2b$. At $x=1$: $f=40\Rightarrow a+b+c=-1$; $f'=2\Rightarrow3a+2b+c=2$; $f''=4\Rightarrow6a+2b=4\Rightarrow3a+b=2$.
To find the values of $a$, $b$, and $c$, we start by using the given information about the function $f(x) = ax^3 + bx^2 + cx + 41$ and its derivatives.
Step 1: Evaluate $f(1)$ to get an equation involving $a$, $b$, and $c$.
Given $f(1) = 40$, we substitute $x = 1$ into $f(x)$:
$$f(1) = a(1)^3 + b(1)^2 + c(1) + 41 = a + b + c + 41 = 40$$
This simplifies to:
$$a + b + c = -1$$
Step 2: Find $f'(x)$ and evaluate $f'(1)$ to get another equation.
The derivative $f'(x)$ is:
$$f'(x) = 3ax^2 + 2bx + c$$
Given $f'(1) = 2$, we substitute $x = 1$ into $f'(x)$:
$$f'(1) = 3a(1)^2 + 2b(1) + c = 3a + 2b + c = 2$$
Step 3: Find $f''(x)$ and evaluate $f''(1)$ to get yet another equation.
The second derivative $f''(x)$ is:
$$f''(x) = 6ax + 2b$$
Given $f''(1) = 4$, we substitute $x = 1$ into $f''(x)$:
$$f''(1) = 6a(1) + 2b = 6a + 2b = 4$$
Step 4: Solve the system of equations obtained from Steps 1, 2, and 3.
We have three equations:
$$\begin{aligned}
a + b + c &= -1 \\
3a + 2b + c &= 2 \\
6a + 2b &= 4
\end{aligned}$$
From the third equation, we can express $b$ in terms of $a$:
$$2b = 4 - 6a \implies b = 2 - 3a$$
Step 5: Substitute $b = 2 - 3a$ into the first and second equations to solve for $a$, $b$, and $c$.
Substituting $b$ into the first equation:
$$a + (2 - 3a) + c = -1 \implies -2a + c = -3$$
And substituting $b$ into the second equation:
$$3a + 2(2 - 3a) + c = 2 \implies 3a + 4 - 6a + c = 2 \implies -3a + c = -2$$
Step 6: Solve for $a$ and $c$ using the equations from Step 5.
We have:
$$\begin{aligned}
-2a + c &= -3 \\
-3a + c &= -2
\end{aligned}$$
Subtracting the second equation from the first gives:
$$a = -1$$
Substituting $a = -1$ back into one of the equations to find $c$:
$$-2(-1) + c = -3 \implies 2 + c = -3 \implies c = -5$$
And using $a = -1$ to find $b$:
$$b = 2 - 3(-1) = 2 + 3 = 5$$
Step 7: Calculate $a^2 + b^2 + c^2$.
Now that we have $a = -1$, $b = 5$, and $c = -5$, we can find $a^2 + b^2 + c^2$:
$$a^2 + b^2 + c^2 = (-1)^2 + 5^2 + (-5)^2 = 1 + 25 + 25 = 51$$
Therefore: $a^2+b^2+c^2$ is equal to option 3, which is 51.
<div class="key-concept"><strong>Key Concept:</strong> $f'(x)=3ax^2+2bx+c$, $f''(x)=6ax+2b$. At $x=1$: $f=40\Rightarrow a+b+c=-1$; $f'=2\Rightarrow3a+2b+c=2$; $f''=4\Rightarrow6a+2b=4\Rightarrow3a+b=2$.</div>
<div class="trap-box"><strong>Trap:</strong> From equations: $a=-1$, $b=5$, $c=-5$. $a^2+b^2+c^2=1+25+25=51$.</div>
Correct Answer: 3