Given:
The equation of the curve is given as,
[tex]f(x)\text{ = 3\lparen2x}^2-4x)^3[/tex]
The required line is the tangent to the given curve at the point (1, -24).
Required:
Equation of the line tangent to the given curve.
Explanation:
Differentiating the given function with respect to x,
[tex]\begin{gathered} f^{\prime}(x)\text{ = 3}\frac{d}{dx}\left(2x^2-4x\right)^3 \\ f^{\prime}(x)=3\times3\times\left(2x^2-4x\right)^2\times\frac{d}{dx}\left(2x^2-4x\right) \\ f^{\prime}(x)=9\left(2x^2-4x\right)^2(4x-4) \end{gathered}[/tex]
Slope of the tangent at (1,-24) is calculated as,
[tex]\begin{gathered} f^{\prime}(x)\text{ at \lparen1,-24\rparen = 9}\times[2\left(1\right)^2-4(1)]^2\times[4(1)-4] \\ f^{\prime}(x)\text{ at \lparen1,-24\rparen = 9}\times[2-4]\times[4-4] \\ f^{\prime}(x)\text{ at \lparen1,-24\rparen = 9}\times-2\times0\text{ } \\ f^{\prime}(x)\text{ at \lparen1,-24\rparen = 0} \\ \end{gathered}[/tex]
Equation of the required line is given as,
[tex]\begin{gathered} (y-(-24))=0(x-1) \\ y+24\text{ = 0} \\ y\text{ = -24} \end{gathered}[/tex]
Answer:
Thus the equation of the required line is
[tex]y=-24[/tex]
