The abscissae of the points,where the tangent | vedclass

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

Question

(D) Given the curve equation: $y = x^3 - 3x^2 - 9x + 5$. To find the slope of the tangent,we differentiate $y$ with respect to $x$: $\frac{dy}{dx} = 3

Vedclass · Accepted Answer

$x = -1$ and $3$

Vedclass · Answer

(D) Given the curve equation: $y = x^3 - 3x^2 - 9x + 5$. To find the slope of the tangent,we differentiate $y$ with respect to $x$: $\frac{dy}{dx} = 3x^2 - 6x - 9$. Since the tangent is parallel to the $x$-axis,its slope must be zero: $\frac{dy}{dx} = 0$. Setting the derivative to zero: $3x^2 - 6x - 9 = 0$. Dividing by $3$: $x^2 - 2x - 3 = 0$. Factoring the quadratic equation: $(x - 3)(x + 1) = 0$. Thus,the abscissae are $x = 3$ and $x = -1$.

The abscissae of the points,where the tangent to the curve $y = x^3 - 3x^2 - 9x + 5$ is parallel to the $x$-axis,are

Similar Questions

Let $f(x) = x^2 e^{-2x}, x > 0$. The maximum value of $f(x)$ is

$A$ wire of length $8 \text{ units}$ is cut into two parts which are bent respectively in the form of a square and a circle. The least value of the sum of the areas so formed is

If $f(x) = 7e^{\sin^2 x} - e^{\cos^2 x} + 2$,then $\sqrt{7f_{\min} + f_{\max}}$ is equal to

The condition that $f(x) = ax^3 + bx^2 + cx + d$ has no extreme value is

If the height of a cone of greatest volume that can be inscribed in a sphere of radius $R$ is $kR$,then the ratio of the volume of the cone to the volume of the sphere is

Vedclass Test Series

Exam Paper Generator

Online Exam Module