Provided f(x) is continuous at x = 1, determine if the function is differentiable at x = 1 using...

Question:

Provided f(x) is continuous at x = 1, determine if the function is differentiable at x = 1 using the limit process.

{eq}f(x) = \left\{\begin{matrix} 3-x, &x<1 \\ \sqrt{4x}, **** x \geq 1 \end{matrix}\right. {/eq}

Function by Differentiability:


By the definition of the differentiability of a function, we'll compute the left-hand side and right-hand side derivatives of the function using the formula shown below:

{eq}\displaystyle f'(x)=\displaystyle\lim_{h \to 0}\frac{f(x+h)-f(x)}{h} {/eq}

According to the condition, the value from the left and right-hand side must be the same.

{eq}\displaystyle\lim_{h \to 0^{-}}\frac{f(x+h)-f(x)}{h}=\displaystyle\lim_{h \to 0^+}\frac{f(x+h)-f(x)}{h} {/eq}

Answer and Explanation:


Given information:

The function for diffferent intervals is:

{eq}f(x) =\left\{\begin{matrix} 3-x, &x<1 \\ \sqrt{4x}, & x \geq 1 \end{matrix}\right. {/eq}

Using the above information about the function, the value of the quotient limit formula from the left-hand side is:

{eq}\begin{align*} \displaystyle\lim_{h \to 0^{-}}\frac{f(x+h)-f(x)}{h}&=\displaystyle\lim_{h \to 0^{-}}\frac{(3-(x+h))-(3-x)}{h}\\ &=\displaystyle\lim_{h \to 0^{-}}\frac{3-x-h-3+x}{h}\\ &=\displaystyle\lim_{h \to 0^{-}}\frac{-h}{h}\\ &=\displaystyle\lim_{h \to 0^{-}}(-1)\\ &=-1 \end{align*} {/eq}


Using the information of the given function, the value of the quotient limit formula from the right-hand side is:

{eq}\begin{align*} \displaystyle\lim_{h \to 0^{+}}\frac{f(x+h)-f(x)}{h}&=\displaystyle\lim_{h \to 0^{+}}\frac{\sqrt{4(x+h)}-\sqrt{4x}}{h}\\ &=\displaystyle\lim_{h \to 0^{+}}\frac{\sqrt{4x+4h}-\sqrt{4x}}{h}\\ &=\displaystyle\lim_{h \to 0^{+}}\frac{\frac{\mathrm{d}(\sqrt{4x+4h}-\sqrt{4x}) }{\mathrm{d} h}}{\frac{\mathrm{d} }{\mathrm{d} h}h}\\ &=\displaystyle\lim_{h \to 0^{+}}\frac{\frac{1}{2\sqrt{4x+4h}}\cdot \frac{\mathrm{d} (4x+4h)}{\mathrm{d} h}-0}{1}\\ &=\displaystyle\lim_{h \to 0^{+}}\frac{\frac{1}{2\sqrt{4x+4h}}\cdot (0+4(1))}{1}\\ &=\displaystyle\lim_{h \to 0^{+}}\frac{2}{\sqrt{4x+4h}}\\ &=\displaystyle \frac{2}{\sqrt{4x+4(0)}}\\ &=\displaystyle \frac{2}{\sqrt{4x}}\\ &=\displaystyle \frac{2}{2\sqrt{x}}\\ &=\displaystyle \frac{1}{\sqrt{x}}\\ \end{align*} {/eq}

Here, we have:

{eq}\displaystyle\lim_{h \to 0^{+}}\frac{f(x+h)-f(x)}{h}\neq \lim_{h \to 0^{-}}\frac{f(x+h)-f(x)}{h} {/eq}

Thus, the given function is differentiable at {eq}x=1 {/eq}.


Learn more about this topic:

Loading...
When to Use the Quotient Rule for Differentiation

from Math 104: Calculus

Chapter 8 / Lesson 8
51K

Related to this Question

Explore our homework questions and answers library