What is an equation of the line tangent to the graph of f(x)=x2(2x+1)3 at the point where x=-1?
(that's x squared (2x+1)cubed)
I'm having trouble finding the derivative:f'(x)which would be the slope (m).
To find the derivative, you must use the product rule
f'(x) = u'*v + u*v'
So,
f(x) = x^2*(2x+1)^3
f'(x)=(2x)(2x+1)^3+(x^2)(3)(2x+1)^2(2)
(need chain rule, too)
Simplify
(2x)(2x+1)^2(5x+1)
(10x^2+2x)(2x+1)^2
In the form
y-y1 = m(x-x1)
y+1 = (10x^2+2x)(2x+1)^2(x+1)
To find the equation of the line tangent to the graph of f(x) at the point where x = -1, we first need to find the slope (m) of the tangent line. The slope can be determined by finding the derivative of the function f(x). Let's go through the steps to find the derivative:
Step 1: Start with the given function f(x) = x^2(2x + 1)^3.
Step 2: Expand the expression inside the parentheses using the power rule: (2x + 1)^3 = 8x^3 + 12x^2 + 6x + 1.
Step 3: Apply the product rule to find the derivative. The product rule states that the derivative of the product of two functions u(x) and v(x) is given by u'(x)v(x) + u(x)v'(x). In this case, the two functions are x^2 and (8x^3 + 12x^2 + 6x + 1).
f'(x) = (x^2)'(8x^3 + 12x^2 + 6x + 1) + (x^2)(8x^3 + 12x^2 + 6x + 1)'.
Step 4: Simplify each derivative using the power rule:
f'(x) = 2x(8x^3 + 12x^2 + 6x + 1) + x^2(24x^2 + 24x + 6).
Step 5: Combine like terms:
f'(x) = 16x^4 + 24x^3 + 12x^2 + 2x + 24x^3 + 24x^2 + 6x^2 + 6x.
Step 6: Simplify further:
f'(x) = 16x^4 + 48x^3 + 18x^2 + 8x.
Now that we have the derivative, we can evaluate it at x = -1 to find the slope at that point.
Step 7: Substitute x = -1 into the derivative:
f'(-1) = 16(-1)^4 + 48(-1)^3 + 18(-1)^2 + 8(-1).
Simplifying, we get:
f'(-1) = 16 - 48 - 18 - 8 = -58.
Therefore, the slope (m) of the tangent line to the graph of f(x) at x = -1 is -58.
To find the equation of the line tangent to the graph at x = -1, we have the slope (-58) and the point (-1, f(-1)). We can use the point-slope form of the equation of a line to get the equation.
Step 8: Plug in the values into the point-slope form:
y - y1 = m(x - x1), where (x1, y1) = (-1, f(-1)).
Substituting these values:
y - f(-1) = -58(x - (-1)).
Simplifying:
y - f(-1) = -58(x + 1).
This is the equation of the line tangent to the graph of f(x) = x^2(2x+1)^3 at the point where x = -1.