Maximum slope of the curve y = -x^3 + 3x^2 + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the function be $f(x) = -x^3 + 3x^2 + 9x - 27$.The slope of the curve at any point $x$ is given by the derivative $f'(x) = -3x^2 + 6x + 9$.To

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) Let the function be $f(x) = -x^3 + 3x^2 + 9x - 27$.The slope of the curve at any point $x$ is given by the derivative $f'(x) = -3x^2 + 6x + 9$.To find the maximum slope,let $g(x) = f'(x) = -3x^2 + 6x + 9$.We find the derivative of $g(x)$ with respect to $x$: $g'(x) = -6x + 6$.Setting $g'(x) = 0$ for critical points gives $-6x + 6 = 0$,which implies $x = 1$.To verify the maximum,we find the second derivative: $g''(x) = -6$. Since $g''(1) = -6 Substituting $x = 1$ into $g(x)$,we get the maximum slope: $g(1) = -3(1)^2 + 6(1) + 9 = -3 + 6 + 9 = 12$.

Maximum slope of the curve $y = -x^3 + 3x^2 + 9x - 27$ is

Similar Questions

The maximum slope of the curve $y = \frac{1}{2} x^{4} - 5 x^{3} + 18 x^{2} - 19 x$ occurs at the point

$A$ cone of maximum volume is inscribed in a given sphere. Then the ratio of the height of the cone to the diameter of the sphere is:

$A$ wire of length $2$ units is cut into two parts,which are bent respectively to form a square of side $x$ units and a circle of radius $r$ units. If the sum of the areas of the square and the circle so formed is minimum,then:

Let $a$ be a real number such that the function $f(x) = ax^2 + 6x - 15, x \in R$ is increasing in $(-\infty, \frac{3}{4})$ and decreasing in $(\frac{3}{4}, \infty)$. Then the function $g(x) = ax^2 - 6x + 15, x \in R$ has a:

The height of the cone of maximum volume inscribed in a sphere of radius $R$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module