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Exercise 8.1 · Q3

Q.Find a linear approximation for the following functions at the indicated points.

(i) f(x)=x3−5x+12, x0=2f(x)=x^3-5x+12,\ x_0=2
(ii) g(x)=x2+9, x0=−4g(x)=\sqrt{x^2+9},\ x_0=-4
(iii) h(x)=xx+1, x0=1h(x)=\dfrac{x}{x+1},\ x_0=1
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✓ Free question

Each part needs f(x0)f(x_0) and f′(x0)f'(x_0) at the given point, then direct substitution into L(x)=f(x0)+f′(x0)(x−x0)L(x)=f(x_0)+f'(x_0)(x-x_0).

Part (i): f(x)=x3−5x+12, x0=2f(x)=x^3-5x+12,\ x_0=2.

f(2)=8−10+12=10f(2)=8-10+12=10. f′(x)=3x2−5f'(x)=3x^2-5, so f′(2)=12−5=7f'(2)=12-5=7.

L(x)=10+7(x−2)=10+7x−14=7x−4L(x)=10+7(x-2)=10+7x-14=7x-4.

Part (ii): g(x)=x2+9, x0=−4g(x)=\sqrt{x^2+9},\ x_0=-4.

g(−4)=16+9=25=5g(-4)=\sqrt{16+9}=\sqrt{25}=5. g′(x)=2x2x2+9=xx2+9g'(x)=\dfrac{2x}{2\sqrt{x^2+9}}=\dfrac{x}{\sqrt{x^2+9}}, so g′(−4)=−45g'(-4)=\dfrac{-4}{5}.

L(x)=5−45(x+4)=5−45x−165=25−165−45x=95−45xL(x)=5-\dfrac45(x+4)=5-\dfrac45x-\dfrac{16}5=\dfrac{25-16}5-\dfrac45x=\dfrac95-\dfrac45x.

Part (iii): h(x)=xx+1, x0=1h(x)=\dfrac{x}{x+1},\ x_0=1.

h(1)=12h(1)=\dfrac12. By the quotient rule, h′(x)=(x+1)(1)−x(1)(x+1)2=1(x+1)2h'(x)=\dfrac{(x+1)(1)-x(1)}{(x+1)^2}=\dfrac{1}{(x+1)^2}, so h′(1)=14h'(1)=\dfrac14.

L(x)=12+14(x−1)=14x+12−14=14x+14=x+14L(x)=\dfrac12+\dfrac14(x-1)=\dfrac14x+\dfrac12-\dfrac14=\dfrac14x+\dfrac14=\dfrac{x+1}{4}.

✓Final answer

(i) L(x)=7x−4L(x)=\boxed{7x-4} (ii) L(x)=−45x+95L(x)=\boxed{-\dfrac45x+\dfrac95} (iii) L(x)=x+14L(x)=\boxed{\dfrac{x+1}{4}}

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