Bonjour Kadoc974
[tex]f(x)=\dfrac{x^2-4x+7}{x^2+3}[/tex]
[tex]1)\ f(-1)=\dfrac{(-1)^2-4\times(-1)+7}{(-1)^2+3}\\\\f(-1)=\dfrac{1+4+7}{1+3}\\\\f(-1)=\dfrac{12}{4} \\\\\boxed{f(-1)=3}\\\\f(1)=\dfrac{1^2-4\times1+7}{1^2+3}\\\\f(1)=\dfrac{1-4+7}{1+3}\\\\f(1)=\dfrac{4}{4} \\\\\boxed{f(1)=1}[/tex]
[tex]2)\ f'(x)=[\dfrac{x^2-4x+7}{x^2+3}]'\\\\f'(x)=\dfrac{(x^2-4x+7)'\times(x^2+3)-(x^2-4x+7)\times(x^2+3)'}{(x^2+3)^2}\\\\f'(x)=\dfrac{(2x-4)(x^2+3)-(x^2-4x+7)\times(2x)}{(x^2+3)^2}\\\\f'(x)=\dfrac{(2x^3+6x-4x^2-12)-(2x^3-8x^2+14x)}{(x^2+3)^2}\\\\f'(x)=\dfrac{2x^3+6x-4x^2-12-2x^3+8x^2-14x}{(x^2+3)^2}\\\\f'(x)=\dfrac{4x^2-8x-12}{(x^2+3)^2}\\\\\boxed{f'(x)=\dfrac{4(x^2-2x-3)}{(x^2+3)^2}}[/tex]
[tex]3)\ f'(-1)=\dfrac{4[(-1)^2-2\times(-1)-3]}{[(-1)^2+3]^2}\\\\f'(-1)=\dfrac{4(1+2-3)}{(1+3)^2}\\\\f'(-1)=\dfrac{4\times0}{16}\\\\\boxed{f'(-1)=0}\\\\f'(1)=\dfrac{4(1^2-2\times1-3)}{(1^2+3)^2}\\\\f'(1)=\dfrac{4(1-2-3)}{(1+3)^2}\\\\f'(1)=\dfrac{4\times(-4)}{16}\\\\f'(1)=\dfrac{-16}{16}\\\\\boxed{f'(1)=-1}[/tex]
[tex]4)\ g(x)=x^2-2x-3\\\\g(x)=0\\\Delta=(-2)^2-4\times1\times(-3)=4+12=16\ \textgreater \ 0\\\\\ x_1=\dfrac{2-\sqrt{16}}{2}=\dfrac{2-4}{2}=\dfrac{-2}{2}=-1\\\\\ x_2=\dfrac{2+\sqrt{16}}{2}=\dfrac{2+4}{2}=\dfrac{6}{2}=3\\\\\\\begin{array}{|c|ccccccc|} x&-\infty&&-1&&3&&+\infty \\ x^2-2x-3&&+&0&-&0&+&\\ \end{array}[/tex]
D'où :
g(x) > 0 <==> x ∈ ]-oo ; -1[ U ]3 ; +oo[
g(x) = 0 <==> x = -1 ou x = 3
g(x) < 0 <==> x ∈ ]-1 ; 3[
[tex]5)\ f'(x)=\dfrac{4(x^2-2x-3)}{(x^2+3)^2}\\\\\\\begin{array}{|c|ccccccc|} x&-\infty&&-1&&3&&+\infty \\ 4(x^2-2x-3)&&+&0&-&0&+&\\(x^2+3)^2&&+&+&+&+&+&\\f'(x)&&+&0&-&0&+&\\ \end{array}[/tex]
[tex]6)\\\\ \begin{array}{|c|ccccccc|} x&-\infty&&-1&&3&&+\infty \\f'(x)&&+&0&-&0&+&\\f(x)&1&\nearrow&3&\searrow&\dfrac{1}{3}&\nearrow&1\\ \end{array}[/tex]
