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Exercise 7.3 · Q4

Q.Using the Lagrange's mean value theorem determine the values of xx at which the tangent is parallel to the secant line at the end points of the given interval:

(i) f(x)=x3−3x+2, x∈[−2,2]f(x)=x^3-3x+2,\ x\in[-2,2]
(ii) f(x)=(x−2)(x−7), x∈[3,11]f(x)=(x-2)(x-7),\ x\in[3,11]
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Compute the secant slope between the two endpoints first, then solve f′(x)f'(x) = that slope, keeping only roots inside the open interval.

Step 1 (i). f(x)=x3−3x+2, [−2,2]f(x)=x^3-3x+2,\ [-2,2].

f(−2)=−8+6+2=0f(-2)=-8+6+2=0; f(2)=8−6+2=4f(2)=8-6+2=4. Secant slope =4−02−(−2)=1=\dfrac{4-0}{2-(-2)}=1.

f′(x)=3x2−3=1⇒x2=43⇒x=±23=±233f'(x)=3x^2-3=1\Rightarrow x^2=\dfrac43\Rightarrow x=\pm\dfrac{2}{\sqrt3}=\pm\dfrac{2\sqrt3}{3}. Both values (≈±1.155\approx\pm1.155) lie in (−2,2)(-2,2). …

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