Question:
Find the slope of the tangent to the curve y = x3 − 3x + 2 at the point whose x-coordinate is 3.
Find the slope of the tangent to the curve y = x3 − 3x + 2 at the point whose x-coordinate is 3.
Updated On: Oct 13, 2023
Hide Solution
Verified By Collegedunia
Solution and Explanation
The given curve is y = x3 − 3x + 2
=\(\frac{dy}{dx}\)=3x2-3
The slope of the tangent to a curve at (x0, y0) is \((\frac{dy}{dx})\bigg] _{ (x_0,y_0)}\).
Hence, the slope of the tangent at the point where the x-coordinate is 3 is given by,
\((\frac{dy}{dx}) \bigg]_{x=3}\)=3x2-3]x=3=3(3)2-3=27-3=24.
Was this answer helpful?
0
0
Top Questions on Applications of Derivatives
- Let \( f: \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \( f''(x)>0 \) for all \( x \in \mathbb{R} \) and \( f'(a-1) = 0 \), where \( a \) is a real number. Let \( g(x) = f(\tan^2 x - 2\tan x + a) \), \( 0<x<\frac{\pi}{2} \).
Consider the following two statements :
(I) \( g \) is increasing in \( (0, \frac{\pi}{4}) \)
(II) \( g \) is decreasing in \( (\frac{\pi}{4}, \frac{\pi}{2}) \)
Then,- JEE Main - 2026
- Mathematics
- Applications of Derivatives
- Find the equation of the tangent to the curve \(y = x^2\) at the point (1,1).
- BITSAT - 2025
- Mathematics
- Applications of Derivatives
- A boat takes 4 hours to travel 40 km downstream and 6 hours to return. The speed of the stream is:
- VITEEE - 2025
- Mathematics
- Applications of Derivatives
- The equation of the tangent to the curve \(y=x^3-2x+1\) at \((1,0)\) is:
- VITEEE - 2025
- Mathematics
- Applications of Derivatives
- A person travels from A to B at 40 km/h and returns at 60 km/h. The average speed for the entire journey is:
- VITEEE - 2025
- Mathematics
- Applications of Derivatives
View More Questions
Questions Asked in CBSE CLASS XII exam
- Find the value of $x$, if \[ \begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ x \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]
- Two point charges of \( -5\,\mu C \) and \( 2\,\mu C \) are located in free space at \( (-4\,\text{cm}, 0) \) and \( (6\,\text{cm}, 0) \) respectively.
(a) Calculate the amount of work done to separate the two charges at infinite distance.
(b) If this system of charges was initially kept in an electric field \[ \vec{E} = \frac{A}{r^2}, \text{ where } A = 8 \times 10^4\, \text{N}\,\text{C}^{-1}\,\text{m}^2, \] calculate the electrostatic potential energy of the system.- CBSE CLASS XII - 2025
- Electrostatics
- 4,000 shares of ₹ 10 each were forfeited for non-payment of second and final call money of ₹ 2 per share. The minimum amount that the company must collect at the time of reissue of these shares will be :
- CBSE CLASS XII - 2025
- Accounting for Share Capital
- If $y = a \cos(\log x) + b \sin(\log x)$, then $x^2y'' + xy'1$ is:
- CBSE CLASS XII - 2025
- Continuity and differentiability
A ladder of fixed length \( h \) is to be placed along the wall such that it is free to move along the height of the wall.
Based upon the above information, answer the following questions:(iii) (b) If the foot of the ladder, whose length is 5 m, is being pulled towards the wall such that the rate of decrease of distance \( y \) is \( 2 \, \text{m/s} \), then at what rate is the height on the wall \( x \) increasing when the foot of the ladder is 3 m away from the wall?
- CBSE CLASS XII - 2025
- Application of derivatives
View More Questions
Concepts Used:
Tangents and Normals
- A tangent at a degree on the curve could be a straight line that touches the curve at that time and whose slope is up to the derivative of the curve at that point. From the definition, you'll be able to deduce the way to realize the equation of the tangent to the curve at any point.
- Given a function y = f(x), the equation of the tangent for this curve at x = x0
- Slope of tangent (at x=x0) m=dy/dx||x=x0
- A normal at a degree on the curve is a line that intersects the curve at that time and is perpendicular to the tangent at that point. If its slope is given by n, and also the slope of the tangent at that point or the value of the derivative at that point is given by m. then we got
m×n = -1
- The normal to a given curve y = f(x) at a point x = x0
- The slope ‘n’ of the normal: As the normal is perpendicular to the tangent, we have: n=-1/m
Diagram Explaining Tangents and Normal:
