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question 22 write the equation of the line tangent to $f(x)=\\frac{1}{x…

Question

question 22
write the equation of the line tangent to $f(x)=\frac{1}{x^5}$ when $x = 1$.
$y=\frac{1}{5}x+\frac{4}{5}$
$y = -5x + 6$
$y = -5x - 4$
$y = 5x - 4$

Explanation:

Step1: Rewrite the function

$f(x) = x^{-5}$

Step2: Find the derivative (slope)

$f'(x) = -5x^{-6} = -\frac{5}{x^6}$

Step3: Calculate slope at $x=1$

$f'(1) = -\frac{5}{1^6} = -5$

Step4: Find $f(1)$

$f(1) = \frac{1}{1^5} = 1$

Step5: Solve for tangent line

Use point-slope form $y - y_1 = m(x - x_1)$:
$y - 1 = -5(x - 1)$
Simplify: $y = -5x + 5 + 1 = -5x + 6$

Answer:

y = -5x + 6