Question

suppose that $f$ is a function defined as $f(x)=-3x^{3}+x^{2}+7$. find an equation for the tangent line drawn to the graph of $f$ at $x = 1$. use exact values.

Explanation:

Step1: Find the derivative of the function

The derivative of \(f(x)=-3x^{3}+x^{2}+7\) is \(f^{\prime}(x)=-9x^{2}+2x\) (using the power rule \((x^{n})^\prime = nx^{n - 1}\)).

Step2: Evaluate the function and its derivative at \(x = 1\)

• Evaluate \(f(1)\): \(f(1)=-3(1)^{3}+(1)^{2}+7=-3 + 1+7=5\)
• Evaluate \(f^{\prime}(1)\): \(f^{\prime}(1)=-9(1)^{2}+2(1)=-9 + 2=-7\)

Step3: Use the point - slope form of a line

The point - slope form is \(y - y_{0}=m(x - x_{0})\), where \((x_{0},y_{0})=(1,5)\) and \(m = f^{\prime}(1)=-7\).
Substituting into the formula: \(y - 5=-7(x - 1)\)
Expand: \(y-5=-7x + 7\)

Step4: Write the equation in slope - intercept form

Add 5 to both sides: \(y=-7x+7 + 5\)

Answer:

\(y=-7x + 12\)