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Exercise 8.2 · Q3

Q.Find Δf\Delta f and dfdf for the function ff for the indicated values of x,Δxx,\Delta x and compare.

(i) f(x)=x3−2x2; x=2, Δx=dx=0.5f(x)=x^3-2x^2;\ x=2,\ \Delta x=dx=0.5
(ii) f(x)=x2+2x+3; x=−0.5, Δx=dx=0.1f(x)=x^2+2x+3;\ x=-0.5,\ \Delta x=dx=0.1
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✓ Free question

Compute the exact change Δf=f(x+Δx)−f(x)\Delta f=f(x+\Delta x)-f(x) directly, and the approximate (tangent-line) change df=f′(x) dxdf=f'(x)\,dx, then compare.

Part (i): f(x)=x3−2x2; x=2, Δx=dx=0.5f(x)=x^3-2x^2;\ x=2,\ \Delta x=dx=0.5.

Step 1. Exact change Δf\Delta f. f(2)=8−8=0f(2)=8-8=0. f(2.5)=15.625−2(6.25)=15.625−12.5=3.125f(2.5)=15.625-2(6.25)=15.625-12.5=3.125. So Δf=f(2.5)−f(2)=3.125−0=3.125\Delta f=f(2.5)-f(2)=3.125-0=3.125.

Step 2. Approximate change dfdf. f(x)=x3−2x2⇒f′(x)=3x2−4xf(x)=x^3-2x^2\Rightarrow f'(x)=3x^2-4x. At x=2x=2: f′(2)=3(4)−4(2)=12−8=4f'(2)=3(4)-4(2)=12-8=4. So df=f′(2) dx=4(0.5)=2df=f'(2)\,dx=4(0.5)=2.

Step 3. Compare. Δf=3.125\Delta f=3.125 and df=2df=2 — close but not equal, since Δx=0.5\Delta x=0.5 is not infinitesimally small; the gap Δf−df=1.125\Delta f-df=1.125 is the higher-order error the linear approximation ignores.

Part (ii): f(x)=x2+2x+3; x=−0.5, Δx=dx=0.1f(x)=x^2+2x+3;\ x=-0.5,\ \Delta x=dx=0.1.

Step 1. Exact change Δf\Delta f. f(−0.5)=0.25−1+3=2.25f(-0.5)=0.25-1+3=2.25. f(−0.4)=0.16−0.8+3=2.36f(-0.4)=0.16-0.8+3=2.36. So Δf=2.36−2.25=0.11\Delta f=2.36-2.25=0.11.

Step 2. Approximate change dfdf. f′(x)=2x+2f'(x)=2x+2, so f′(−0.5)=−1+2=1f'(-0.5)=-1+2=1. df=f′(−0.5) dx=1(0.1)=0.1df=f'(-0.5)\,dx=1(0.1)=0.1.

Step 3. Compare. Δf=0.11\Delta f=0.11 and df=0.1df=0.1 — again close, with a small gap of 0.010.01 from the finite (not infinitesimal) Δx\Delta x.

✓Final answer

(i) Δf=3.125, df=2\Delta f=\boxed{3.125},\ df=\boxed{2} (ii) Δf=0.11, df=0.1\Delta f=\boxed{0.11},\ df=\boxed{0.1} — both cases confirm Δf≈df\Delta f\approx df.

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