Answered

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Let g be the function given by g(x) = x^4-3^3-x. What are all values of x such that g'(x)=1/2

Sagot :

Answer:

The answer is "[tex]x= \frac{\sqrt[3]{3} }{2}[/tex]"

Step-by-step explanation:

Given:

[tex]\to g(x) = x^4-3^3-x \\\\\to g'(x) = \frac{1}{2}\\\\x=?[/tex]

Solve:

[tex]\to g(x) = x^4-3^3-x\\\\\to g(x) =x^4 -27-x \\\\\to g(x) =x^4 -x-27[/tex]

differentiate the above equation:

[tex]\to g'(x) = 4x^3- 1[/tex]

Given that:

[tex]\to g'(x)= \frac{1}{2}[/tex]

[tex]\to \frac{1}{2}= 4x^3 -1\\\\\to \frac{1}{2} +1= 4x^3 \\\\\to \frac{1+2}{2}= 4x^3 \\\\\to \frac{3}{2}= 4x^3 \\\\\to x^3 =\frac{3}{2 \times 4} \\\\\to x^3 =\frac{3}{8} \\\\ \to \bold{x= \frac{\sqrt[3]{3} }{2}}[/tex]

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