Name: Derivative at a Point
Uploaded: 2019-07-19
Duration: 5 min 23 s
Description: How will you define the continuity of a function? Let us learn about the relationship between continuity and differentiation of a function at a point. Watch on!

Question

Find the values of

a

and

b

so that the function

f (x) = \{\begin{cases} x^{2} + 3 x + a, & if x \leq 1 \\ b x + 2, & if x > 1 \end{cases}

is differentiable at each

x \in R

.

Accepted Answer

Given: $f (x) = \{\begin{cases} x^{2} + 3 x + a, & if x \leq 1 \\ b x + 2, & if x > 1 \end{cases}$

It is given that the function is differentiable at each $x \in R$ and every differentiable function is continuous.

So $f (x)$ is continuous at $x = 1$

Therefore, $\lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x) = f (1)$

$\Rightarrow \lim_{x \to 1} (x^{2} + 3 x + a) = \lim_{x \to 1} (b x + 2) = a + 4$

$\Rightarrow a + 4 = b + 2 = a + 4 \to (1)$

Since, $f (x)$ is differentiable at $x = 1$ .

So, $(LHD at x = 1) = (RHD at x = 1)$

Thus, $\lim_{x \to 1^{-}} \frac{f (x) - f (1)}{x - 1} = \lim_{x \to 1^{+}} \frac{f (x) - f (1)}{x - 1}$

$\Rightarrow \lim_{x \to 1} \frac{x^{2} + 3 x + a - a - 4}{x - 1} = \lim_{x \to 1} \frac{b x + 2 - 4 - a}{x - 1}$

$\Rightarrow \lim_{x \to 1} \frac{x^{2} + 3 x - 4}{x - 1} = \lim_{x \to 1} \frac{b x + 2 - 4 - a}{x - 1}$

$\Rightarrow \lim_{x \to 1} \frac{x^{2} + 4 x - x - 4}{x - 1} = \lim_{x \to 1} \frac{b x + 2 - 4 - a}{x - 1}$

$\Rightarrow \lim_{x \to 1} \frac{(x + 4) (x - 1)}{x - 1} = \lim_{x \to 1} \frac{b x - 2 - a}{x - 1}$

$\Rightarrow \lim_{x \to 1} \frac{(x + 4) (x - 1)}{x - 1} = \lim_{x \to 1} \frac{b x - b}{x - 1}$ [Using $(1)$ ]

$\Rightarrow \lim_{x \to 1} (x + 4) = \lim_{x \to 1} (b)$

$\Rightarrow 5 = b$

From $(1)$ , we have: $a + 4 = b + 2$

$\Rightarrow a + 4 = 5 + 2$

$\Rightarrow a = 7 - 4$

$\Rightarrow a = 3$

Hence, $a = 3, b = 5$ .

Find the values of $a$ and $b$ so that the function $f (x) = \{\begin{cases} x^{2} + 3 x + a, & if x \leq 1 \\ b x + 2, & if x > 1 \end{cases}$ is differentiable at each $x \in R$ .

Important Questions on Differentiability

Learn and Prepare for any exam you want !

Find the values of a and b so that the function fx=x2+3x+a,if x≤1bx+2,if x>1 is differentiable at each x∈R.

Important Questions on Differentiability

Learn and Prepare for any exam you want !

Find the values of $a$ and $b$ so that the function $f (x) = \{\begin{cases} x^{2} + 3 x + a, & if x \leq 1 \\ b x + 2, & if x > 1 \end{cases}$ is differentiable at each $x \in R$ .