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f^(')(x)=-(4)/(x^(2)) and f(2)=4. f(1)=

f′(x)=−4x2 f^{\prime}(x)=-\frac{4}{x^{2}} and f(2)=4 f(2)=4 .\newlinef(1)= f(1)=

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Q. f′(x)=−4x2 f^{\prime}(x)=-\frac{4}{x^{2}} and f(2)=4 f(2)=4 .\newlinef(1)= f(1)=
  1. Find Derivative of f(x)f(x): We know the derivative of f(x)f(x), which is f′(x)=−4x2f'(x) = -\frac{4}{x^{2}}. We need to find f(1)f(1).
  2. Use Derivative to Find Rate of Change: Since we have f(2)=4f(2) = 4, we can use the derivative to find the rate of change of the function, but we can't directly find f(1)f(1) from f′(x)f'(x) without integrating.
  3. Integrate f′(x)f'(x) to Find f(x)f(x): Let's integrate f′(x)f'(x) to get f(x)f(x). The integral of f′(x)=−4x2f'(x) = -\frac{4}{x^{2}} is f(x)=4x+Cf(x) = \frac{4}{x} + C, where CC is the constant of integration.
  4. Find Constant C: We can find the constant C by using the given point f(2)=4f(2) = 4. Plugging in the values, we get 4=42+C4 = \frac{4}{2} + C, which simplifies to 4=2+C4 = 2 + C.
  5. Calculate f(1)f(1): Solving for CC, we get C=4−2C = 4 - 2, which is C=2C = 2.
  6. Calculate f(1)f(1): Solving for CC, we get C=4−2C = 4 - 2, which is C=2C = 2.Now we have the function f(x)=4x+2f(x) = \frac{4}{x} + 2. We can find f(1)f(1) by plugging in x=1x = 1.
  7. Calculate f(1)f(1): Solving for CC, we get C=4−2C = 4 - 2, which is C=2C = 2.Now we have the function f(x)=4x+2f(x) = \frac{4}{x} + 2. We can find f(1)f(1) by plugging in x=1x = 1.f(1)=41+2f(1) = \frac{4}{1} + 2, which simplifies to f(1)=4+2f(1) = 4 + 2.
  8. Calculate f(1)f(1): Solving for CC, we get C=4−2C = 4 - 2, which is C=2C = 2.Now we have the function f(x)=4x+2f(x) = \frac{4}{x} + 2. We can find f(1)f(1) by plugging in x=1x = 1.f(1)=41+2f(1) = \frac{4}{1} + 2, which simplifies to f(1)=4+2f(1) = 4 + 2.So, f(1)=6f(1) = 6.

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