Question

find an equation for the line tangent to the graph of f at (1,7), where f is given by f(x)=3x^3 - 3x^2 + 7. y=

Explanation:

Step1: Find the derivative of $f(x)$

Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, we have $f^\prime(x)=\frac{d}{dx}(3x^{3}-3x^{2}+7)=3\times3x^{2}-3\times2x+0 = 9x^{2}-6x$.

Step2: Evaluate the derivative at $x = 1$

Substitute $x = 1$ into $f^\prime(x)$: $f^\prime(1)=9(1)^{2}-6(1)=9 - 6=3$. The value of $f^\prime(1)$ is the slope $m$ of the tangent line.

Step3: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(1,7)$ and $m = 3$. Substituting these values, we get $y - 7=3(x - 1)$.

Step4: Simplify the equation

Expand the right - hand side: $y-7 = 3x-3$. Then, add 7 to both sides to get $y=3x + 4$.

Answer:

$y=3x + 4$