Question

given the function $f(x)=x^{3}-3x^{2}-4x + 1$, determine all coordinate points $(x,y)$ on the graph of $f$ such that the line tangent to $f$ at $(x,y)$ has a slope of 5. answer attempt 1 out of 2 two solutions and submit answer

Explanation:

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

The derivative of $f(x)=x^{3}-3x^{2}-4x + 1$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ is $f'(x)=3x^{2}-6x-4$.

Step2: Set the derivative equal to the slope

Since the slope of the tangent line is 5, we set $f'(x)=5$. So, $3x^{2}-6x - 4=5$.

Step3: Rearrange the equation

Rearrange $3x^{2}-6x - 4 = 5$ to the standard quadratic form $ax^{2}+bx + c = 0$. We get $3x^{2}-6x-9 = 0$. Divide through by 3 to simplify: $x^{2}-2x - 3=0$.

Step4: Factor the quadratic equation

Factor $x^{2}-2x - 3$ as $(x - 3)(x+1)=0$.

Step5: Solve for $x$

Set each factor equal to zero: $x - 3=0$ gives $x = 3$ and $x+1=0$ gives $x=-1$.

Step6: Find the corresponding $y$ - values

When $x = 3$, $y=f(3)=3^{3}-3\times3^{2}-4\times3 + 1=27-27-12 + 1=-11$.
When $x=-1$, $y=f(-1)=(-1)^{3}-3\times(-1)^{2}-4\times(-1)+1=-1 - 3 + 4+1=1$.

Answer:

$(-1,1)$ and $(3,-11)$