For the function f(x) = 1/x^2 find an expression for the slope of a tangent at the point where a = 1 using . Simplify the expression first before substitution.
using the limits formula....
Well, when you say "find an expression for the slope of a tangent," I can't help but think of skateboards rolling downhill, making people scream "wheeeee!" But let's get serious for a second and tackle this math problem.
To find the slope of a tangent line at the point where a = 1 for the function f(x) = 1/x^2, we need to calculate the derivative of f(x) with respect to x.
So, let's go ahead and differentiate f(x) using the power rule. The power rule states that if we have a function of the form f(x) = x^n, then its derivative is given by f'(x) = n*x^(n-1).
In our case, f(x) = 1/x^2, so using the power rule, we get:
f'(x) = -2/x^(2+1) = -2/x^3.
Now, we want to find the slope of the tangent line at a = 1, so let's substitute a = 1 into the derivative expression:
f'(1) = -2/1^3 = -2.
And there you have it! The expression for the slope of the tangent line at the point where a = 1 for the function f(x) = 1/x^2 is simply -2.
Now, let's take a moment to appreciate how awesome skateboarding clowns would be. Picture a clown doing tricks on a skateboard while juggling flaming torches. That's the kind of slope I'd like to see!
To find the slope of the tangent line at the point where x = a, we can use the derivative of the function. The function f(x) = 1/x^2 can be rewritten as f(x) = x^(-2).
Taking the derivative of f(x) with respect to x, we use the power rule for differentiation:
f'(x) = -2x^(-3).
Now, to find the slope of the tangent line at x = 1 (let a = 1), we substitute x = 1 into the derivative:
f'(1) = -2(1)^(-3).
Simplifying the expression:
1^(-3) = 1, since any number raised to the power of -3 is equal to its reciprocal cubed.
Therefore, f'(1) = -2.
The slope of the tangent line at x = 1 is -2.
To find the slope of a tangent line to the curve represented by the function f(x) = 1/x^2 at the point where x = 1, we can use the derivative. The derivative of f(x) with respect to x, denoted as f'(x), gives us the slope of the tangent line at any given point.
Let's start by finding the derivative of f(x).
Step 1: Find the derivative of f(x) = 1/x^2
The power rule of differentiation states that if we have a function in the form f(x) = x^n, where n is a constant, the derivative is given by f'(x) = nx^(n-1).
Applying this rule, we can find the derivative of 1/x^2 as follows:
f'(x) = d/dx (1/x^2)
= -2/x^3
Step 2: Simplify the expression
Now, we need to simplify the expression obtained for the derivative.
To simplify -2/x^3, we can use the properties of exponents.
Recall that negative exponents can be converted into positive exponents by moving the term to the denominator of a fraction.
Simplifying -2/x^3:
-2/x^3 = -2 * x^-3
= -2/x^3
Hence, the expression for the derivative of f(x) = 1/x^2, after simplification, is f'(x) = -2/x^3.
Step 3: Substitution
Now that we have the simplified expression, we can substitute x = 1 into the derivative function to find the slope of the tangent line at that point.
Substituting x = 1 into f'(x) = -2/x^3:
f'(1) = -2/(1^3)
= -2/1
= -2
Therefore, the slope of the tangent line at the point where x = 1 for the function f(x) = 1/x^2 is -2.
point A
f(x) = 1/(x^2)
point B
f(x+deltaX) = 1/((x+deltaX)^2)
The slope of the line between these points is:
[f(x+deltaX) - f(x)]/[(x+deltaX - x]
The slope of the tangent is given when deltaX->0.
to get you started...
[(1/(x+deltaX)^2) - 1/x^2]/[x+deltaX - x]
Get rid of the fractions in the numerator and reduce the fraction as much as possible. Then take the limit as deltaX->0.