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Exercise 7.3 · Q2

Q.Using the Rolle's theorem, determine the values of xx at which the tangent is parallel to the xx-axis for the following functions:

(i) f(x)=x2−x, x∈[0,1]f(x)=x^2-x,\ x\in[0,1]
(ii) f(x)=x2−2xx+2, x∈[−1,6]f(x)=\dfrac{x^2-2x}{x+2},\ x\in[-1,6]
(iii) f(x)=x−x3, x∈[0,9]f(x)=\sqrt x-\dfrac{x}{3},\ x\in[0,9]
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✓ Free question

In each part, verify f(a)=f(b)f(a)=f(b) (Rolle's hypothesis), then solve f′(x)=0f'(x)=0 and discard any root falling outside the open interval.

Step 1 (i). f(x)=x2−x, [0,1]f(x)=x^2-x,\ [0,1].

f(0)=0=f(1)f(0)=0=f(1). f′(x)=2x−1=0⇒x=12∈(0,1)f'(x)=2x-1=0\Rightarrow x=\dfrac12\in(0,1). ✓

Step 2 (ii). f(x)=x2−2xx+2, [−1,6]f(x)=\dfrac{x^2-2x}{x+2},\ [-1,6].

f(−1)=1+21=3f(-1)=\dfrac{1+2}{1}=3; f(6)=36−128=3f(6)=\dfrac{36-12}{8}=3. Equal, so Rolle's applies.

f′(x)=(2x−2)(x+2)−(x2−2x)(1)(x+2)2=x2+4x−4(x+2)2.f'(x)=\frac{(2x-2)(x+2)-(x^2-2x)(1)}{(x+2)^2}=\frac{x^2+4x-4}{(x+2)^2}.

Set the numerator to 00: x2+4x−4=0⇒x=−4±322=−2±22x^2+4x-4=0\Rightarrow x=\dfrac{-4\pm\sqrt{32}}{2}=-2\pm2\sqrt2.

−2+22≈0.83∈(−1,6)-2+2\sqrt2\approx0.83\in(-1,6) ✓;  −2−22≈−4.83∉(−1,6)\ -2-2\sqrt2\approx-4.83\notin(-1,6) (discard).

Step 3 (iii). f(x)=x−x3, [0,9]f(x)=\sqrt x-\dfrac x3,\ [0,9].

f(0)=0f(0)=0; f(9)=3−3=0f(9)=3-3=0. Equal.

f′(x)=12x−13=0⇒12x=13⇒x=32⇒x=94∈(0,9). ✓f'(x)=\frac{1}{2\sqrt x}-\frac13=0\Rightarrow \frac{1}{2\sqrt x}=\frac13\Rightarrow\sqrt x=\frac32\Rightarrow x=\frac94\in(0,9).\ ✓

✓Final answer

(i) x=12x=\dfrac12. (ii) x=22−2≈0.83x=2\sqrt2-2\approx0.83. (iii) x=94x=\dfrac94.

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