Question:
For the curve \( y = 3x^3 - 3x^2 + 1 \) at \( x = 1 \), find the equation of the tangent.
For the curve \( y = 3x^3 - 3x^2 + 1 \) at \( x = 1 \), find the equation of the tangent.
Show Hint
To find the tangent equation quickly:
1. Differentiate the function to get the slope.
2. Substitute the given \(x\) value to find the slope \(m\).
3. Find the corresponding \(y\)-coordinate.
4. Use the point–slope formula \(y - y_1 = m(x - x_1)\).
Updated On: Apr 15, 2026
- \( y = 3x - 2 \)
- \( y = 3x - 3 \)
- \( y = 3x - 1 \)
- \( y = x + 1 \)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Concept:
The equation of the tangent to a curve \(y = f(x)\) at \(x = a\) is given by:
\[
y - f(a) = f'(a)(x - a)
\]
where:
Step 1: {Differentiate the given function.} Given \[ y = 3x^3 - 3x^2 + 1 \] Differentiating with respect to \(x\): \[ \frac{dy}{dx} = 9x^2 - 6x \] This represents the slope of the tangent at any point \(x\).
Step 2: {Find the slope at \(x = 1\).} \[ m = 9(1)^2 - 6(1) \] \[ m = 9 - 6 = 3 \] Thus, the slope of the tangent is \(m = 3\).
Step 3: {Find the point on the curve at \(x = 1\).} Substitute \(x=1\) into the original equation: \[ y = 3(1)^3 - 3(1)^2 + 1 \] \[ y = 3 - 3 + 1 \] \[ y = 1 \] So the point of tangency is \[ (1,1) \]
Step 4: {Use the point–slope form of the tangent line.} \[ y - y_1 = m(x - x_1) \] Substituting \(m=3\), \(x_1=1\), \(y_1=1\): \[ y - 1 = 3(x - 1) \] \[ y - 1 = 3x - 3 \] \[ y = 3x - 2 \]
- \(f'(a)\) is the slope of the tangent obtained using differentiation.
- \(f(a)\) is the point on the curve where the tangent touches.
Step 1: {Differentiate the given function.} Given \[ y = 3x^3 - 3x^2 + 1 \] Differentiating with respect to \(x\): \[ \frac{dy}{dx} = 9x^2 - 6x \] This represents the slope of the tangent at any point \(x\).
Step 2: {Find the slope at \(x = 1\).} \[ m = 9(1)^2 - 6(1) \] \[ m = 9 - 6 = 3 \] Thus, the slope of the tangent is \(m = 3\).
Step 3: {Find the point on the curve at \(x = 1\).} Substitute \(x=1\) into the original equation: \[ y = 3(1)^3 - 3(1)^2 + 1 \] \[ y = 3 - 3 + 1 \] \[ y = 1 \] So the point of tangency is \[ (1,1) \]
Step 4: {Use the point–slope form of the tangent line.} \[ y - y_1 = m(x - x_1) \] Substituting \(m=3\), \(x_1=1\), \(y_1=1\): \[ y - 1 = 3(x - 1) \] \[ y - 1 = 3x - 3 \] \[ y = 3x - 2 \]
Was this answer helpful?
0
0
Top MHT CET Mathematics Questions
- The line $5x + y - 1 = 0$ coincides with one of the lines given by $5x^2 + xy - kx - 2y + 2 = 0 $ then the value of k is
- If $\int\frac{f\left(x\right)}{log \left(sin\,x\right)}dx = log\left[log\,sin\,x\right]+c$ then $f\left(x\right)=$
- If $\int\limits^{K}_0 \frac{dx}{2 + 18 x^2} = \frac{\pi}{24}$, then the value of K is
- If $\int^{\pi/2}_{0} \log\cos x dx =\frac{\pi}{2} \log\left(\frac{1}{2}\right)$ then $ \int^{\pi/2}_{0} \log\sec x dx = $
- The point on the curve $y = \sqrt{x - 1}$ where the tangent is perpendicular to the line $2x + y - 5 = 0 $ is
View More Questions
Top MHT CET Tangents and Normals Questions
- The point on the curve $y = \sqrt{x - 1}$ where the tangent is perpendicular to the line $2x + y - 5 = 0 $ is
- The value of \( \int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} dx \) is equal to:
- Let the curve be represented by $x=2(\cos t+t\sin t)$, $y=2(\sin t-t\cos t)$. Then normal at any point '$t$' of the curve is at a distance of _____ units from the origin.
Top MHT CET Questions
- During r- DNA technology, which one of the following enzymes is used for cleaving DNA molecule ?
- In non uniform circular motion, the ratio of tangential to radial acceleration is (r = radius of circle, $v =$ speed of the particle, $\alpha =$ angular acceleration)
- The temperature of $32^{\circ}C$ is equivalent to
- The line $5x + y - 1 = 0$ coincides with one of the lines given by $5x^2 + xy - kx - 2y + 2 = 0 $ then the value of k is
- The heat of formation of water is $ 260\, kJ $ . How much $ H_2O $ is decomposed by $ 130\, kJ $ of heat ?
View More Questions