hw 9 - the tangent line section 2.4: problem 3 (1 point)\nlet $f(x)=5x^{3}-x^{2}-7x + 7$.(a) find…

hw 9 - the tangent line section 2.4: problem 3 (1 point)\nlet $f(x)=5x^{3}-x^{2}-7x + 7$.(a) find $f(x)$.$f(x)=square$(b) find the slope of the line tangent to the graph of $f$ at $x = 6$.slope at $x = 6:square$ (use at least 2 decimal places)(c) find an equation of the line tangent to the graph of $f$ at $x = 6$.tangent line: $y=square$

hw 9 - the tangent line section 2.4: problem 3 (1 point)\nlet $f(x)=5x^{3}-x^{2}-7x + 7$.(a) find $f(x)$.$f(x)=square$(b) find the slope of the line tangent to the graph of $f$ at $x = 6$.slope at $x = 6:square$ (use at least 2 decimal places)(c) find an equation of the line tangent to the graph of $f$ at $x = 6$.tangent line: $y=square$

Answer

Explanation:

Step1: Differentiate (f(x))

Using the power - rule ((x^n)^\prime=nx^{n - 1}), for (f(x)=5x^{3}-x^{2}-7x + 7), we have (f^\prime(x)=5\times3x^{2}-2x-7=15x^{2}-2x - 7).

Step2: Find the slope at (x = 6)

Substitute (x = 6) into (f^\prime(x)). (f^\prime(6)=15\times6^{2}-2\times6 - 7=15\times36-12 - 7=540-12 - 7 = 521).

Step3: Find the point on the function at (x = 6)

Find (f(6)): (f(6)=5\times6^{3}-6^{2}-7\times6 + 7=5\times216-36 - 42 + 7=1080-36 - 42 + 7=1009).

Step4: Find the equation of the tangent line

Use the point - slope form (y - y_1=m(x - x_1)), where (m = 521), (x_1 = 6), and (y_1 = 1009). (y-1009=521(x - 6)) (y-1009=521x-3126) (y=521x-3126 + 1009) (y=521x-2117)

Answer:

(a) (15x^{2}-2x - 7) (b) (521.00) (c) (521x-2117)