Find the derivative of f(x) = (x -...

MathematicsDifferentiability of a FunctionJEE Advanced 1979Easy

Find the derivative of $f (x) = {\frac{x - 1}{2 x ^{2} - 7 x + 5} - \frac{1}{3} when x \neq = 1 when x = 1$ at $x = 1$ .

Answer:

- 2/9

Visualized Solution (English)

Visualizing the Piecewise Function

Function: $f (x) = {\frac{x - 1}{2 x ^{2} - 7 x + 5} - \frac{1}{3} x \neq = 1 x = 1$
Goal: Find $f^{'} (1)$

Factorizing the Denominator

Denominator: $2 x^{2} - 7 x + 5$
Splitting the middle term: $2 x^{2} - 2 x - 5 x + 5$
Factoring: $2 x (x - 1) - 5 (x - 1) = (x - 1) (2 x - 5)$

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f (x) for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x \neq = 1

For $x \neq = 1$ : $f (x) = \frac{x - 1}{( x - 1 ) ( 2 x - 5 )}$
Simplified form: $f (x) = \frac{1}{2 x - 5}$

The Derivative Definition

Definition: $f^{'} (1) = lim_{h \to 0} \frac{f ( 1 + h ) - f ( 1 )}{h}$
Given: $f (1) = - \frac{1}{3}$

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f (1 + h)

Substitute $x = 1 + h$ into simplified $f (x)$ :
$f (1 + h) = \frac{1}{2 ( 1 + h ) - 5} = \frac{1}{2 h - 3}$

Setting up the Limit

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

Simplifying the Numerator

Numerator: $\frac{3 + ( 2 h - 3 )}{3 ( 2 h - 3 )} = \frac{2 h}{3 ( 2 h - 3 )}$
Full expression: $f^{'} (1) = lim_{h \to 0} \frac{2 h}{3 h ( 2 h - 3 )}$

Cancelling the Indeterminate Form

Cancelling $h$ : $f^{'} (1) = lim_{h \to 0} \frac{2}{3 ( 2 h - 3 )}$

Final Evaluation

Substitute $h = 0$ : $f^{'} (1) = \frac{2}{3 ( 2 ( 0 ) - 3 )}$
$f^{'} (1) = \frac{2}{3 ( - 3 )} = - \frac{2}{9}$

Summary and Key Takeaways

Key Takeaway: Always use the limit definition for piecewise functions at junction points.
Final Answer: $f^{'} (1) = - \frac{2}{9}$
Next Challenge: Check the differentiability if $f (1)$ was changed to $0$ .

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Answer:0, 1

Find the derivative of $f (x) = {\frac{x - 1}{2 x ^{2} - 7 x + 5} - \frac{1}{3} when x \neq = 1 when x = 1$ at $x = 1$ .

Visualized Solution (English)

Visualizing the Piecewise Function

Factorizing the Denominator

Simplifying $f (x)$ for $x \neq = 1$

The Derivative Definition

Calculating $f (1 + h)$

Setting up the Limit

Simplifying the Numerator

Cancelling the Indeterminate Form

Final Evaluation

Summary and Key Takeaways

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the derivative of f(x)={2x2−7x+5x−1​−31​​when x=1when x=1​ at x=1.

Visualized Solution (English)

Visualizing the Piecewise Function

Factorizing the Denominator

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1

The Derivative Definition

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(1+h)

Setting up the Limit

Simplifying the Numerator

Cancelling the Indeterminate Form

Final Evaluation

Summary and Key Takeaways

Conceptually Similar Problems

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