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f^(')(x)=8x^(3)-12 x and f(1)=1. f(-2)=

f′(x)=8x3−12x f^{\prime}(x)=8 x^{3}-12 x and f(1)=1 f(1)=1 .\newlinef(−2)= f(-2)=

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Full solution

Q. f′(x)=8x3−12x f^{\prime}(x)=8 x^{3}-12 x and f(1)=1 f(1)=1 .\newlinef(−2)= f(-2)=
  1. Substitute x=−2x=-2: Substitute −2-2 for xx in the derivative f′(x)f'(x) to find the slope at x=−2x=-2.f′(−2)=8(−2)3−12(−2)f'(-2) = 8(-2)^{3} - 12(-2)
  2. Calculate f′(x)f'(x): Calculate the value of f′(−2)f'(-2).
    f′(−2)=8(−8)−12(2)f'(-2) = 8(-8) - 12(2)
    f′(−2)=−64−24f'(-2) = -64 - 24
    f′(−2)=−88f'(-2) = -88
  3. Need original function f(x)f(x): We have the slope at x=−2x=-2, but we need the actual function value f(−2)f(-2). Since we know f(1)=1f(1)=1, we can't directly use the derivative to find f(−2)f(-2). We need the original function f(x)f(x), which we don't have. We can't proceed without it.

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