Answer :
ANSWER :
a. The derivative is :
[tex]f^{\prime}(x)=12x^{2}-\frac{6}{x^{3}}[/tex]b. The point on the graph is (1, 4)
EXPLANATION :
From the problem, we have the function :
[tex]f(x)=4x^3+\frac{3}{x^2}-3[/tex]a. The derivative will be :
[tex]\begin{gathered} f^{\prime}(x)=3(4x^2)+(-2)(\frac{3}{x^3}) \\ f^{\prime}(x)=12x^2-\frac{6}{x^3} \end{gathered}[/tex]b. The gradient of the tangent is also the derivative which is f'(x)
So we need to find the point in which f'(x) = 6
That will be :
[tex]\begin{gathered} 6=12x^2-\frac{6}{x^3} \\ \text{ Multiply both sides by x\textasciicircum3} \\ 6x^3=12x^5-6 \\ \text{ Divide both sides by 6} \\ x^3=2x^5-1 \\ x^3-2x^5=-1 \\ 2x^5-x^3=1 \\ x^3(2x^2-1)=1 \end{gathered}[/tex]Equate both factors to 1 :
[tex]\begin{gathered} x^3=1 \\ x=1 \\ \\ 2x^2-1=1 \\ 2x^2=2 \\ x^2=1 \\ x=1 \end{gathered}[/tex]So we have x = 1
Substitute x = 1 to the original function :
[tex]\begin{gathered} f(1)=4(1)^3+\frac{3}{(1)^2}-3 \\ f(1)=4+3-3 \\ f(1)=4 \end{gathered}[/tex]Therefore the point is (1, 4)