Function Differentiability Conditions

For a function $f(x)$ to be continuously differentiable at a point $x=c$, it must satisfy two conditions:

• Continuity: The limit from the left must equal the limit from the right, and both must equal the function's value at $c$. Mathematically, $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$.
• Differentiability: The derivative from the left must equal the derivative from the right at $c$. Mathematically, $f'(c^-) = f'(c^+)$.

In this problem, the point of interest is $x=1$. The function is defined as:

$f(x) = \begin{cases} bx+a & \text{if } x < 1, \\ 5x^2 & \text{if } x\geq1. \end{cases}$

Continuity at $x=1$

Applying the continuity condition at $x=1$:

• Limit from the left: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (bx+a) = b(1) + a = b+a$.
• Value at $x=1$ (and limit from the right): $f(1) = 5(1)^2 = 5$.

Equating these gives the first equation:

$b+a = 5 \quad (\text{Equation } 1)$

Derivatives at $x=1$

First, find the derivatives for each piece of the function:

• For $x < 1$, $f'(x) = \frac{d}{dx}(bx+a) = b$.
• For $x > 1$, $f'(x) = \frac{d}{dx}(5x^2) = 10x$.

Now, apply the differentiability condition ($f'(1^-) = f'(1^+)$):

• Derivative from the left: $f'(1^-) = b$.
• Derivative from the right: $f'(1^+) = 10(1) = 10$.

Equating these gives the second equation:

$b = 10 \quad (\text{Equation } 2)$

Solving for Constants $a$ and $b$

We have a system of two linear equations:

1. $b+a = 5$
2. $b = 10$

Substitute the value of $b$ from Equation 2 into Equation 1:

$(10) + a = 5$

Solving for $a$:

$a = 5 - 10 = -5$

So, $a = -5$ and $b = 10$.

Final Calculation of $a^2b$

We need to find the value of $a^2b$. Substitute the values of $a$ and $b$ we found:

$a^2b = (-5)^2 \times (10)$

$a^2b = 25 \times 10$

$a^2b = 250$