If `f(x)={{:(e^(2x^(3+x)),x gt 0),(ax+b, x le 0):}` is differentiable at x = 0, then

Options

A. `a=1, b=-1`
B. `a=-1, b=1`
C. `a=1, b=1`
D. `a=-1, b=-1`

Answer

To determine the values of \( a \) and \( b \) such that the function

\[ f(x) = \begin{cases} e^{2x^{3+x}}, & x > 0 \\ ax + b, & x \leq 0 \end{cases} \]

is differentiable at \( x = 0 \), we need to ensure that both the function is continuous at \( x = 0 \) and that the left-hand derivative equals the right-hand derivative at that point.

### Step 1: Check Continuity at \( x = 0 \)

For \( f(x) \) to be continuous at \( x = 0 \), we need:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \]

Calculating \( f(0) \):

\[ f(0) = a(0) + b = b \]

Calculating the right-hand limit as \( x \to 0^+ \):

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} e^{2x^{3+x}} = e^{2(0)^{3+0}} = e^0 = 1 \]

Now, for continuity:

\[ b = 1 \]

### Step 2: Check Differentiability at \( x = 0 \)

Next, we need to ensure that the left-hand derivative equals the right-hand derivative at \( x = 0 \).

**Left-hand derivative**:

\[ f'(0^-) = \lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^-} \frac{a(h) + b - b}{h} = \lim_{h \to 0^-} \frac{ah}{h} = a \]

**Right-hand derivative**:

\[ f'(0^+) = \lim_{h \to 0^+} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^+} \frac{e^{2h^{3+h}} - 1}{h} \]

Using L'Hôpital's Rule since this is an indeterminate form \( \frac{0}{0} \):

\[ \text{Differentiate the numerator and denominator:} \]

Numerator:

\[ \frac{d}{dh}(e^{2h^{3+h}}) = e^{2h^{3+h}} \cdot \frac{d}{dh}(2h^{3+h}) = e^{2h^{3+h}} \cdot (6h^{2+h} + 2h^{3+h} \ln(h)) \]

Denominator:

\[ \frac{d}{dh}(h) = 1 \]

Thus,

\[ f'(0^+) = \lim_{h \to 0^+} e^{2h^{3+h}} (6h^{2+h} + 2h^{3+h} \ln(h)) \]

As \( h \to 0 \), \( e^{2h^{3+h}} \to 1 \) and \( 6h^{2+h} \to 0 \). Therefore, we can conclude that:

\[ f'(0^+) = 0 \]

### Step 3: Set Left-hand and Right-hand Derivatives Equal

Setting the left-hand derivative equal to the right-hand derivative:

\[ a = 0 \]

### Conclusion

From the continuity condition, we found \( b = 1 \), and from the differentiability condition, we found \( a = 0 \). Therefore, the values of \( a \) and \( b \) are:

\[ a = 0, \quad b = 1 \]