Do question A ,B,C

Question

Do question A ,B,C

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Here are the solutions for questions A, B, and C.

A) If $f(x) = 3x^2 - 2x + 1$ , find $f'(x)$ from first principles.

Write down the definition of the derivative from first principles and find $f(x+h)$ .
The definition of the derivative is:
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
Given $f(x) = 3x^2 - 2x + 1$ , we find $f(x+h)$ :
$f(x+h) = 3(x+h)^2 - 2(x+h) + 1$ $f(x+h) = 3(x^2 + 2xh + h^2) - 2x - 2h + 1$ $f(x+h) = 3x^2 + 6xh + 3h^2 - 2x - 2h + 1$
Calculate $f(x+h) - f(x)$ .
$f(x+h) - f(x) = (3x^2 + 6xh + 3h^2 - 2x - 2h + 1) - (3x^2 - 2x + 1)$ $f(x+h) - f(x) = 3x^2 + 6xh + 3h^2 - 2x - 2h + 1 - 3x^2 + 2x - 1$ $f(x+h) - f(x) = 6xh + 3h^2 - 2h$
Divide by $h$ and simplify.
$\frac{f(x+h) - f(x)}{h} = \frac{6xh + 3h^2 - 2h}{h}$ $\frac{f(x+h) - f(x)}{h} = \frac{h(6x + 3h - 2)}{h}$ $\frac{f(x+h) - f(x)}{h} = 6x + 3h - 2$
Take the limit as $h \to 0$ .
$f'(x) = \lim_{h \to 0} (6x + 3h - 2)$ $f'(x) = 6x + 3(0) - 2$ $\boxed{\mathbf{f'(x) = 6x - 2}}$
B) Differentiate $y = \frac{3x^2 - 2x + 1}{\sqrt{x}}$ with respect to $x$ .
Rewrite the function by expressing $\sqrt{x}$ as $x^{1/2}$ and dividing each term in the numerator.
$y = \frac{3x^2}{x^{1/2}} - \frac{2x}{x^{1/2}} + \frac{1}{x^{1/2}}$ $y = 3x^{2 - 1/2} - 2x^{1 - 1/2} + x^{-1/2}$ $y = 3x^{3/2} - 2x^{1/2} + x^{-1/2}$
Differentiate each term using the power rule, $\frac{d}{dx}(ax^n) = anx^{n-1}$ .
$\frac{dy}{dx} = \frac{d}{dx}(3x^{3/2}) - \frac{d}{dx}(2x^{1/2}) + \frac{d}{dx}(x^{-1/2})$ $\frac{dy}{dx} = 3 \left(\frac{3}{2}\right) x^{\frac{3}{2} - 1} - 2 \left(\frac{1}{2}\right) x^{\frac{1}{2} - 1} + \left(-\frac{1}{2}\right) x^{-\frac{1}{2} - 1}$ $\frac{dy}{dx} = \frac{9}{2} x^{1/2} - 1 x^{-1/2} - \frac{1}{2} x^{-3/2}$
Simplify the expression.
$\boxed{\mathbf{\frac{dy}{dx} = \frac{9}{2} \sqrt{x} - \frac{1}{\sqrt{x}} - \frac{1}{2x\sqrt{x}}}}$
C) Find the equation of the tangent to the curve $y = x^3 - 2x^2 + 3x - 1$ at the point where $x=2$ .
Find the y-coordinate of the point of tangency.
Substitute $x=2$ into the equation of the curve:
$y = (2)^3 - 2(2)^2 + 3(2) - 1$ $y = 8 - 2(4) + 6 - 1$ $y = 8 - 8 + 6 - 1$ $y = 5$
The point of tangency is $(2, 5)$ .
Find the derivative $\frac{dy}{dx}$ to determine the slope function.
$y = x^3 - 2x^2 + 3x - 1$ $\frac{dy}{dx} = 3x^2 - 2(2x) + 3(1) - 0$ $\frac{dy}{dx} = 3x^2 - 4x + 3$
Evaluate the derivative at $x=2$ to find the slope ( $m$ ) of the tangent.
$m = 3(2)^2 - 4(2) + 3$ $m = 3(4) - 8 + 3$ $m = 12 - 8 + 3$ $m = 7$
Use the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$ , to find the equation of the tangent.
Using the point $(x_1, y_1) = (2, 5)$ and slope $m=7$ :
$y - 5 = 7(x - 2)$ $y - 5 = 7x - 14$ $y = 7x - 14 + 5$ $\boxed{\mathbf{y = 7x - 9}}$

Do question A ,B,C

Step-by-step solution

Write down the definition of the derivative from first principles and find $f(x+h)$ .

Calculate $f(x+h) - f(x)$ .

Divide by $h$ and simplify.

Take the limit as $h \to 0$ .

Rewrite the function by expressing $\sqrt{x}$ as $x^{1/2}$ and dividing each term in the numerator.

Differentiate each term using the power rule, $\frac{d}{dx}(ax^n) = anx^{n-1}$ .

Simplify the expression.

Find the y-coordinate of the point of tangency.

Find the derivative $\frac{dy}{dx}$ to determine the slope function.

Evaluate the derivative at $x=2$ to find the slope ( $m$ ) of the tangent.

Use the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$ , to find the equation of the tangent.

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More Mathematics Questions

Do question A ,B,C

Step-by-step solution

Write down the definition of the derivative from first principles and find f(x+h)f(x+h)f(x+h).

Calculate f(x+h)−f(x)f(x+h) - f(x)f(x+h)−f(x).

Divide by hhh and simplify.

Take the limit as h→0h \to 0h→0.

Rewrite the function by expressing x\sqrt{x}x​ as x1/2x^{1/2}x1/2 and dividing each term in the numerator.

Differentiate each term using the power rule, ddx(axn)=anxn−1\frac{d}{dx}(ax^n) = anx^{n-1}dxd​(axn)=anxn−1.

Simplify the expression.

Find the y-coordinate of the point of tangency.

Find the derivative dydx\frac{dy}{dx}dxdy​ to determine the slope function.

Evaluate the derivative at x=2x=2x=2 to find the slope (mmm) of the tangent.

Use the point-slope form of a linear equation, y−y1=m(x−x1)y - y_1 = m(x - x_1)y−y1​=m(x−x1​), to find the equation of the tangent.

Still stuck on this one?

More Mathematics Questions

Write down the definition of the derivative from first principles and find $f(x+h)$ .

Calculate $f(x+h) - f(x)$ .

Divide by $h$ and simplify.

Take the limit as $h \to 0$ .

Rewrite the function by expressing $\sqrt{x}$ as $x^{1/2}$ and dividing each term in the numerator.

Differentiate each term using the power rule, $\frac{d}{dx}(ax^n) = anx^{n-1}$ .

Find the derivative $\frac{dy}{dx}$ to determine the slope function.

Evaluate the derivative at $x=2$ to find the slope ( $m$ ) of the tangent.

Use the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$ , to find the equation of the tangent.