Find an equation of the tangent line to the graph of f(x) = (x3
− 3x2 + 1)(2x4
− 1)
at x = −1.
f(x) = (x^3 − 3x^2 + 1)(2x^4 − 1)
using the product rule, ...
f ' (x) = (x^3 - 3x^2 + 1)(8x) + (2x^4 - 1)(3x^2 - 6x)
so f ' (-1) = (-1 - 3 + 1)(-8) + (2 - 1)(3 + 6)
= 24 + 9 = 33
f(-1) = (+1-3+1)(2-1) = -3
so you have the slope of 33 and a point (-1,-3)
y+3 = 33(x+1)
y = 33x + 30
thankyou!
To find the equation of the tangent line to the graph of a function, we need to find the slope of the tangent line and a point on the line.
First, let's find the slope of the tangent line at x = -1. We can do this by taking the derivative of the function f(x) with respect to x.
f(x) = (x^3 - 3x^2 + 1)(2x^4 - 1)
To simplify the function, expand the product:
f(x) = 2x^7 - x^3 - 6x^6 + 3x^4 + 2x^4 - 1
Now, take the derivative of f(x) with respect to x:
f'(x) = 14x^6 - 3x^2 - 36x^5 + 12x^3 + 8x^3
Next, we evaluate the derivative at x = -1:
f'(-1) = 14(-1)^6 - 3(-1)^2 - 36(-1)^5 + 12(-1)^3 + 8(-1)^3
= 14(1) - 3(1) - 36(1) + 12(-1) + 8(-1)
= 14 - 3 - 36 - 12 - 8
= -45
Therefore, the slope of the tangent line at x = -1 is -45.
Now, let's find a point on the line. We can use the original function f(x) to find the y-coordinate at x = -1.
f(-1) = (-1^3 - 3(-1)^2 + 1)(2(-1)^4 - 1)
= (-1 + 3 + 1)(2 - 1)
= (3)(1)
= 3
So, we have a point (-1, 3) on the tangent line.
Finally, we can use the slope-intercept form of a line to write the equation of the tangent line:
y - y1 = m(x - x1)
where (x1, y1) is a point on the line and m is the slope.
Plugging in the values, we have:
y - 3 = -45(x - (-1))
Simplifying,
y - 3 = -45(x + 1)
y - 3 = -45x - 45
y = -45x - 42
Therefore, the equation of the tangent line to the graph of f(x) = (x^3 - 3x^2 + 1)(2x^4 - 1) at x = -1 is y = -45x - 42.