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Help me out as quickly as you can, timed assignment!

Help me out as quickly as you can, timed assignment!-example-1
User Patriotic
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\begin{gathered} f(x)=3x^3-7x-5 \\ f^(\prime)(x)=\lim _(h\to0)(f(x+h)-f(x))/(h) \\ f^(\prime)(x)=\lim _(h\to0)(3(x+h)^3-7(x+h)-5-(3x^3-7x-5))/(h) \\ (x+h)^3=x^3+3x^2h+3xh^2+h^3 \\ f^(\prime)(x)=\lim _(h\to0)\frac{3(x^3+3x^2h+3xh^2+h^3)^{}-7(x+h)-5-(3x^3-7x-5)}{h} \\ \\ f^(\prime)(x)=\lim _(h\to0)(3x^3+9x^2h+9xh^2+3h^3-7x-7h-5-3x^3+7x+5)/(h) \\ f^(\prime)(x)=\lim _(h\to0)((3x^3-3x^3)+9x^2h+9xh^2+3h^3+(-7x+7x)-7h+(-5+5))/(h) \\ f^(\prime)(x)=\lim _(h\to0)(9x^2h+9xh^2+3h^3-7h)/(h) \\ f^(\prime)(x)=\lim _(h\to0)\frac{h(9x^2+9xh^{}+3h^2-7)}{h} \\ f^(\prime)(x)=\lim _(h\to0)9x^2+9xh^{}+3h^2-7=9x^2+9x(0)+3(0)^2-7=9x^2-7 \\ f^(\prime)(x)=9x^2-7 \\ The\text{ first derivative is }9x^2-7 \\ For\text{ f''(x)} \\ f^(\prime\prime)(x)=\lim _(h\to0)(f^(\prime)(x+h)-f^(\prime)(x))/(h) \\ f^(\prime\prime)(x)=\lim _(h\to0)(9(x+h)^2-7-(9x^2-7))/(h) \\ (x+h)^2=x^2+2xh+h^2 \\ f^(\prime\prime)(x)=\lim _(h\to0)\frac{9(x^2+2xh+h^2)^{}-7-(9x^2-7)}{h} \\ f^(\prime\prime)(x)=\lim _(h\to0)\frac{9x^2+18xh+9h^2^{}-7-9x^2+7}{h} \\ f^(\prime\prime)(x)=\lim _(h\to0)((9x^2-9x^2)+18xh+9h^2+(-7+7))/(h) \\ f^(\prime\prime)(x)=\lim _(h\to0)(18xh+9h^2)/(h) \\ f^(\prime\prime)(x)=\lim _(h\to0)(h(18x+9h))/(h) \\ f^(\prime\prime)(x)=\lim _(h\to0)18x+9h=18x+9(0)=18x \\ f^(\prime\prime)(x)=18x \\ \text{The second derivative is 18x} \end{gathered}

User Sehafoc
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