Use linear approximation to approximate f(0.8) given f(x)=2x^3-3x+1
I'm not sure how to do it without a point given.
f(1) = 2-3+1 = 0
df = (6x^2-3) dx
So, starting from f(1),
f(0.8) = f(1)+(6-3)(-0.2) = -1.8
The line is
y-0 = 3(x-1)
You can see the approximation at
http://www.wolframalpha.com/input/?i=plot+2x^3-3x%2B1%2C3%28x-1%29
You could have started with a different value of x, say, x=0.9, and used a smaller dx, but x=1 is convenient and moderately close.
To use linear approximation to approximate a value of a function, you can use the equation of the tangent line to the function at a given point. In this case, we need a point on the graph of f(x)=2x^3-3x+1 to find the tangent line.
Let's choose the point (a, f(a)), where a = 1. Since f(x)=2x^3-3x+1, we can evaluate f(1) as follows:
f(a) = 2a^3 - 3a + 1
f(1) = 2(1)^3 - 3(1) + 1
f(1) = 2 - 3 + 1
f(1) = 0
Therefore, the point on the graph of f(x)=2x^3-3x+1 that we will use is (1, 0).
Now, we will find the slope of the tangent line at (1, 0) by taking the derivative of the function f(x). The derivative of f(x)=2x^3-3x+1 is:
f'(x) = 6x^2 - 3
Substituting x=1 into f'(x), we get:
f'(1) = 6(1)^2 - 3
f'(1) = 6 - 3
f'(1) = 3
So, the slope of the tangent line at (1, 0) is 3.
Now, we can find the equation of the tangent line using the point-slope form:
y - y1 = m(x - x1)
where (x1, y1) is the point (1, 0) and m is the slope.
Substituting the values, we have:
y - 0 = 3(x - 1)
y = 3x - 3
This is the equation of the tangent line to the function f(x)=2x^3-3x+1 at (1, 0).
Using this tangent line, we can approximate f(0.8) by substituting x=0.8 into the equation of the tangent line:
y = 3(0.8) - 3
Calculating this, we get:
y ≈ -0.6
Therefore, using linear approximation, we can approximate f(0.8) to be approximately -0.6.
To use linear approximation, you first need to find the equation of a tangent line to the function at a given point. In this case, we need to find the equation of the tangent line to f(x) = 2x^3 - 3x + 1 at x = 0.8.
Step 1: Find the derivative of f(x).
Take the derivative of the function f(x) = 2x^3 - 3x + 1 to find the slope of the tangent line at any given point.
f'(x) = 6x^2 - 3
Step 2: Find the slope of the tangent line at x = 0.8.
Substitute x = 0.8 into the derivative f'(x) to find the slope of the tangent line at that point.
f'(0.8) = 6(0.8)^2 - 3 = 3.84 - 3 = 0.84
Step 3: Use the point-slope form of a linear equation.
The equation of a line can be expressed using the point-slope form: y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is the slope of the line.
We are given that x₁ = 0.8 and f(0.8) = y₁. So, we have:
y - f(0.8) = 0.84(x - 0.8)
Step 4: Substitute f(0.8) into the equation.
Substitute the value of f(0.8) into the equation to find the equation of the tangent line.
y - f(0.8) = 0.84(x - 0.8)
y - f(0.8) = 0.84x - 0.672
Step 5: Simplify and rearrange the equation.
Rearrange the equation in the form y = mx + b, where m is the slope and b is the y-intercept.
y = 0.84x - 0.672 + f(0.8)
Step 6: Substitute x = 0.8 into the equation.
Substitute x = 0.8 into the equation to find the value of f(0.8) using linear approximation.
f(0.8) ≈ 0.84(0.8) - 0.672 + f(0.8)
Step 7: Solve for f(0.8).
Simplify the equation by canceling out f(0.8) on both sides and solve for f(0.8).
0 ≈ 0.84(0.8) - 0.672
0 ≈ 0.672 - 0.672
0 ≈ 0
Therefore, f(0.8) is approximately equal to 0 using linear approximation.