Question

Compute f'and find an equation of the tangent line to the graph at x=a.

f(x)=3x⁻³,a=5. (Use symbolic notation and fractions where needed. Let y=f(x) and express equation in terms of x and y.)
f'(x)= _____
equation:____

Answer 1

The derivative of the function f(x) = 3x⁻³ is f'(x)= -9x⁻⁴. At x=5, the slope of the tangent is -9/625, and thus the equation of the tangent line is y - 3/125 = (-9/625)(x - 5).

Step-by-step explanation:

To compute f'(x) and find an equation of the tangent line to the graph at x=a, we start with the given function f(x) = 3x⁻³. The first step is to find the derivative of the function to calculate the slope of the tangent line at any point x.

The power rule for derivatives states that if f(x) = x^n, then f'(x) = nx^(n-1). Applying this rule to our function, we get:

f'(x) = d/dx(3x⁻³) = 3 × (-3)x^(-3-1) = -9x⁻⁴

At x=a, which in our case is x=5, we can calculate the slope of the tangent line:

f'(5) = -9(5)⁻⁴ = -9/(5⁴) = -9/625

Now, the equation of the tangent line at x=a in point-slope form is y - f(a) = f'(a)(x - a). Substituting a=5 and f(a) = 3(5)⁻³ = 3/125, we get:

y - 3/125 = (-9/625)(x - 5)

Hence, f'(x)= -9x⁻⁴, and the equation of the tangent line at x=5 is y - 3/125 = (-9/625)(x - 5).