Question

Calculate the gradient of the curve y=x^3 -6x^2+3x-1 at point x= -1,..Find also the minimum gradient of the curve..

Answer 1

gradient means slope of curve .
here,
y = x³ - 6x² + 3x - 1
$\frac{dy}{dx}_{x=-1}=3x^{3-1}-6\times\:2x^{2-1}+3\\\\\frac{dy}{dx}_{x=-1}=3x^2-12x+3\\\\gradient=3(-1)^2-12(-1)+3=3+12+3=18$
now, Let dy/dx = g(x)
g(x) = 3x² - 12x + 3
differentiate with respect to x
g'(x) = 6x - 12 = 0 => x = 2
again, differentiate with respect to x
g"(x) = 6 > 0 hence, at x = 2 , g(x) gain minimum value.
hence, minimum value of g(x) = 3(2)²-12(2)+3=-9

hence, minimum value of gradient = -9

Answer 2

gradient means slope of curve .

here,

y = x³ - 6x² + 3x - 1

$\begin{lgathered}\frac{dy}{dx}_{x=-1}=3x^{3-1}-6\times\:2x^{2-1}+3\\\\\frac{dy}{dx}_{x=-1}=3x^2-12x+3\\\\gradient=3(-1)^2-12(-1)+3=3+12+3=18\end{lgathered} \:$

now, Let dy/dx = g(x)

g(x) = 3x² - 12x + 3

differentiate with respect to x

g'(x) = 6x - 12 = 0 => x = 2

again, differentiate with respect to x

g"(x) = 6 > 0 hence, at x = 2 , g(x) gain minimum value.

hence, minimum value of g(x) = 3(2)²-12(2)+3=-9

hence, minimum value of gradient = -9