If f(x) = begin{cases} A + Bx^2, & x

Question

(A) For $f(x)$ to be differentiable at $x = 1$,it must first be continuous at $x = 1$.$\lim_{x 	o 1^-} f(x) = f(1)$$A + B(1)^2 = 3A(1) - B + 2$$A + B

Vedclass · Accepted Answer

$2, 3$

Vedclass · Answer

(A) For $f(x)$ to be differentiable at $x = 1$,it must first be continuous at $x = 1$.$\lim_{x 	o 1^-} f(x) = f(1)$$A + B(1)^2 = 3A(1) - B + 2$$A + B = 3A - B + 2$$2B = 2A + 2 \implies B = A + 1$ (Equation $1$)Now,for differentiability,the left-hand derivative must equal the right-hand derivative at $x = 1$.$f'(1^+) = \frac{d}{dx}(3Ax - B + 2) = 3A$$f'(1^-) = \frac{d}{dx}(A + Bx^2) = 2Bx$At $x = 1$,$f'(1^-) = 2B(1) = 2B$.Equating the derivatives: $3A = 2B$ (Equation $2$)Substitute Equation $1$ into Equation $2$:$3A = 2(A + 1)$$3A = 2A + 2$$A = 2$Using Equation $1$,$B = 2 + 1 = 3$.Thus,$A = 2$ and $B = 3$.

If $f(x) = \begin{cases} A + Bx^2, & x < 1 \\ 3Ax - B + 2, & x \geqslant 1 \end{cases}$,then find $A$ and $B$ so that $f(x)$ is differentiable at $x = 1$.

Similar Questions

If the function $g(x)=\begin{cases} K \sqrt{x+1} &, 0 \leq x \leq 3 \\ mx+2 &, 3 < x \leq 5 \end{cases}$ is differentiable, then $K+m=$

Let $f:R \to R$ be a function defined by $f(x) = \text{Min}\{x + 1, |x| + 1\}$. Then which of the following is true?

Let $f:R \to R$ be a function defined by $f(x) = \max \,(x, x^3).$ The set of all points where $f(x)$ is not differentiable is

Let $S = \{t \in R \mid f(x) = |x - \pi|(e^{|x|} - 1) \sin |x| \text{ is not differentiable at } t\}$. Then $S$ is

$f(x) = \begin{cases} 4, & -\infty < x < -\sqrt{5} \\ x^2-1, & -\sqrt{5} \leq x \leq \sqrt{5} \\ 4, & \sqrt{5} < x < \infty \end{cases}$
If $k$ is the number of points where $f(x)$ is not differentiable,then $k-2=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module