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Question Number 199538 by cherokeesay last updated on 04/Nov/23

Answered by mr W last updated on 05/Nov/23

$$a)$$$$f\left(x\right)=x^3-1-m(x-1)$$$$f^{\prime }\left(x\right)=3x^2-m=0$$$$x^3-1-3x^2(x-1)=0$$$$2x^3-3x^2+1=0$$$${(x-1)}^2(2x+1)=0$$$$\Rightarrow x=1\Rightarrow m=3$$$$\Rightarrow x=-\frac{1}{2}\Rightarrow m=\frac{3}{4}$$$$b)$$$$m=3$$$$f\left(x\right)=x^3-1-3(x-1)={(x-1)}^2(x+2)$$$$zerosatx=-2,x=1$$$$...$$$$c)$$$${\int }_{-2}^1(x^3-3x+2)dx$$$$=\frac{1}{4}(1^4-2^4)-\frac{3}{2}(1^2-2^2)+2(1+2)=\frac{27}{4}$$

Commented by mr W last updated on 05/Nov/23

Commented by cherokeesay last updated on 05/Nov/23

$$thankyousomuchmaster!!!$$$$$$

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