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Exercise 11.4 · Q3

Q.If f′′(x)=12x−6f''(x) = 12x-6 and f(1)=30, f′(1)=5f(1)=30,\ f'(1)=5, find f(x)f(x).

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Since only the second derivative is given, integrate once to get f′(x)f'(x) (using f′(1)=5f'(1)=5 to fix that constant), then integrate again to get f(x)f(x) (using f(1)=30f(1)=30 to fix the second constant).

Step 1. Integrate f′′(x)f''(x) once to get f′(x)f'(x).

f′(x)=∫(12x−6) dx=6x2−6x+c1.f'(x) = \int (12x-6)\,dx = 6x^2-6x+c_1.

Step 2. Apply f′(1)=5f'(1)=5 to find c1c_1.

f′(1)=6(1)2−6(1)+c1=6−6+c1=c1.f'(1) = 6(1)^2-6(1)+c_1 = 6-6+c_1=c_1.

Setting this equal to 55: c1=5c_1=5, so f′(x)=6x2−6x+5f'(x)=6x^2-6x+5.

Step 3. Integrate f′(x)f'(x) to get f(x)f(x).

f(x)=∫(6x2−6x+5) dx=2x3−3x2+5x+c2.f(x) = \int (6x^2-6x+5)\,dx = 2x^3-3x^2+5x+c_2.

Step 4. Apply f(1)=30f(1)=30 to find c2c_2. …

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