Consider the function f(x) = 2x
2
+ 1. Find the equation of the tangent to the
graph of f(x) at x = 2 (when calculating f'(2, use first principles)
To find the equation of the tangent to the graph of the function f(x) = 2x^2 + 1 at x = 2 using first principles, you need to follow these steps:
Step 1: Calculate the derivative of the function f(x) to find the slope (gradient) of the tangent line. We use the first principles approach to do this.
The derivative of f(x) is denoted by f'(x) and is given by the following formula:
f'(x) = lim(h->0) [f(x + h) - f(x)] / h
Substituting the function f(x) = 2x^2 + 1 into the above formula, we get:
f'(x) = lim(h->0) [2(x + h)^2 + 1 - (2x^2 + 1)] / h
Step 2: Simplify the expression by expanding and collecting like terms within the limit:
f'(x) = lim(h->0) [2(x^2 + 2xh + h^2) + 1 - 2x^2 - 1] / h
= lim(h->0) [2x^2 + 4xh + 2h^2 + 1 - 2x^2 - 1] / h
= lim(h->0) [4xh + 2h^2] / h
= lim(h->0) 4x + 2h
Step 3: Calculate the limit as h approaches 0:
f'(x) = 4x + lim(h->0) 2h
= 4x
Therefore, the derivative (slope) of the function f(x) at any point is given by f'(x) = 4x.
Step 4: Substitute the value of x = 2 into the derivative equation to find the slope at x = 2:
f'(2) = 4(2)
= 8
So, at x = 2, the slope of the tangent line is 8.
Step 5: Use the point-slope formula to determine the equation of the tangent line. The point-slope formula is given by:
y - y1 = m(x - x1)
Where (x1, y1) is a point on the line and m is the slope.
Using the point (2, f(2)) = (2, 2(2)^2 + 1) = (2, 9) and the slope m = 8, we have:
y - 9 = 8(x - 2)
Expanding this equation, we get the equation of the tangent line to the graph of f(x) at x = 2:
y - 9 = 8x - 16
Simplifying further, we obtain:
y = 8x - 7
Therefore, the equation of the tangent line to the graph of f(x) = 2x^2 + 1 at x = 2 is y = 8x - 7.