How do you find the tangent line of #f(x) = 32x^2 # at x=1?
Take the derivative and plug in the desired
We can calculate the slope from the derivative of the function:
Now that we have our slope, we can evaluate the original function and solve for the yintercept:
Now, we know the slope and intercept, and can write the equation for the tangent line:
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To find the tangent line of f(x) = 32x^2 at x=1, we need to find the derivative of the function and evaluate it at x=1. The derivative of f(x) is f'(x) = 4x. Evaluating f'(1), we get f'(1) = 4(1) = 4. Therefore, the slope of the tangent line at x=1 is 4. To find the equation of the tangent line, we use the pointslope form y  y1 = m(x  x1), where (x1, y1) is the point on the curve. Plugging in x=1, y=f(1), and m=4, we have y  (32(1)^2) = 4(x  (1)). Simplifying this equation gives us the equation of the tangent line as y = 4x + 7.
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To find the tangent line of ( f(x) = 3  2x^2 ) at ( x = 1 ), follow these steps:
 Find the derivative of ( f(x) ) using the power rule: ( f'(x) = 4x ).
 Evaluate the derivative at ( x = 1 ): ( f'(1) = 4(1) = 4 ).
 Use the pointslope form of a line, ( y  y_1 = m(x  x_1) ), where ( m ) is the slope and ( (x_1, y_1) ) is a point on the line.
 Substitute the values ( x_1 = 1 ) and ( m = 4 ) into the pointslope form.
 Simplify the equation to obtain the tangent line equation.
Therefore, the tangent line of ( f(x) = 3  2x^2 ) at ( x = 1 ) is ( y = 7  4x ).
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To find the tangent line of (f(x) = 3  2x^2) at (x = 1), follow these steps:
 Find the derivative of (f(x)) with respect to (x) to get the slope of the tangent line.
 Evaluate the derivative at (x = 1) to find the slope of the tangent line at that point.
 Use the pointslope form of the equation of a line, (y  y_1 = m(x  x_1)), where (m) is the slope of the tangent line and ((x_1, y_1)) is the point of tangency, which is ((1, f(1))).
 Plug in the values of (m) and ((x_1, y_1)) into the pointslope form to get the equation of the tangent line.
Let's go through the steps:

Find the derivative of (f(x)): [f'(x) = \frac{d}{dx}(3  2x^2) = 4x]

Evaluate the derivative at (x = 1): [f'(1) = 4(1) = 4]

Determine the point of tangency: [f(1) = 3  2(1)^2 = 3  2 = 1] So, the point of tangency is ((1, 1)).

Use the pointslope form: [y  1 = 4(x + 1)]

Simplify the equation: [y  1 = 4x + 4] [y = 4x + 5]
Therefore, the equation of the tangent line to (f(x) = 3  2x^2) at (x = 1) is (y = 4x + 5).
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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