Question

(1) Test whether the function f(x) = (2x + 3)³/² is differentiable at x = -3/2
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Answer 1

Given,

f(x)= $(2x+3)^{\frac{3}{2} }$                              and x=  $\frac{-3}{2}$

Now, we know that a function is differentiable at a point =a if and only if

Right hand derivative(RHD) = Left hand derivative(LHD) at that point.

So,

RHD= $\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$

      = $\lim_{h \to \ 0} \frac{f(\frac{-3}{2} +h)-f(\frac{-3}{2} )}{h}$

     = $\lim_{h \to \ 0} \frac{(2h)^{\frac{3}{2} } - 0}{h}$

     = $\lim_{h \to \ 0} 2^{\frac{3}{2} } \sqrt{h}$

     = 0

and

LHD= $\lim_{h \to 0} \frac{f(a-h)-f(a)}{-h}$

      = $\lim_{h \to \ 0} \frac{f(\frac{-3}{2} -h)-f(\frac{-3}{2} )}{-h}$

      = $\lim_{h \to \ 0} \frac{(-2h)^{\frac{3}{2} } - 0}{-h}$

      = $\lim_{h \to \ 0} -(-2)^{\frac{3}{2} } \sqrt{h}$

      = 0.

Since, RHD= LHD

then the given function is differentiable at x=  $\frac{-3}{2}$.

Answer 2

Answer:

Yes, the function $f(x)=(2x+3)^{3/2}$ is differentiable at $x=\frac{-3}{2}$.

Step-by-step explanation:

Consider the function as follows:

$f(x)=(2x+3)^{3/2}$

Differentiate with respect to $x$.

$f'(x)=\frac{3}{2}(2x+3)^{1/2}.2$

$f'(x)=3(2x+3)^{1/2}$ ...... (1)

Substitute the value $\frac{-3}{2}$ for $x$ in the equation (1) as follows:

$f'(\frac{-3}{2} )=3(2(\frac{-3}{2} )+3)^{1/2}$

$f'(\frac{-3}{2} )=3(-3 +3)^{1/2}$

$f'(\frac{-3}{2} )=3(0)^{1/2}$

$f'(\frac{-3}{2} )=0$

The value of $f(x)$ at $x=\frac{-3}{2}$ exists and is finite.

Therefore, the function $f(x)=(2x+3)^{3/2}$ is differentiable at $x=\frac{-3}{2}$.

#SPJ3