Math Snap

STEP 1

Assumptions
1. The function for the curve is $y = x^3 - 3x - 5$.
2. We need to find the gradient of the curve at $x = 2$.
3. We need to find the coordinates of the maximum and minimum turning points of the curve.
4. For part (b), we need to derive the function $y = 2x^2 + 4x - 3$ from first principles.
5. For part (c), we need to use integration by substitution to evaluate the integral $\int_{0}^{1} (3x^2 + 2)^4 \, dx$.
6. For part (d), we need to use the product rule to find $\frac{dy}{dx}$ for the given functions.

STEP 2

To find the gradient of the curve at $x = 2$, we first need to find the derivative of the function $y = x^3 - 3x - 5$.
$$\frac{dy}{dx} = \frac{d}{dx}(x^3 - 3x - 5)$$

STEP 3

Differentiate the function term by term.
$$\frac{dy}{dx} = 3x^2 - 3$$

STEP 4

Evaluate the derivative at $x = 2$.
$$\left. \frac{dy}{dx} \right|_{x=2} = 3(2)^2 - 3$$

STEP 5

Calculate the value.
$$\left. \frac{dy}{dx} \right|_{x=2} = 3(4) - 3 = 12 - 3 = 9$$ The gradient of the curve at $x = 2$ is 9.

STEP 6

To find the coordinates of the maximum and minimum turning points, we need to find the critical points by setting the derivative equal to zero.
$$3x^2 - 3 = 0$$

STEP 7

Solve for $x$.
$$3x^2 = 3$$ $$x^2 = 1$$ $$x = \pm 1$$

STEP 8

Determine the y-coordinates by substituting $x = 1$ and $x = -1$ into the original function $y = x^3 - 3x - 5$.
For $x = 1$:
$$y = 1^3 - 3(1) - 5 = 1 - 3 - 5 = -7$$ For $x = -1$:
$$y = (-1)^3 - 3(-1) - 5 = -1 + 3 - 5 = -3$$

STEP 9

The coordinates of the turning points are $(1, -7)$ and $(-1, -3)$.

STEP 10

To determine whether these points are maxima or minima, we need to find the second derivative of the function.
$$\frac{d^2y}{dx^2} = \frac{d}{dx}(3x^2 - 3)$$

STEP 11

Differentiate the first derivative.
$$\frac{d^2y}{dx^2} = 6x$$

STEP 12

Evaluate the second derivative at $x = 1$ and $x = -1$.
For $x = 1$:
$$\left. \frac{d^2y}{dx^2} \right|_{x=1} = 6(1) = 6$$ For $x = -1$:
$$\left. \frac{d^2y}{dx^2} \right|_{x=-1} = 6(-1) = -6$$

STEP 13

Since $\frac{d^2y}{dx^2}$ is positive at $x = 1$, the point $(1, -7)$ is a minimum. Since $\frac{d^2y}{dx^2}$ is negative at $x = -1$, the point $(-1, -3)$ is a maximum.

STEP 14

For part (b), we need to find the derivative of $y = 2x^2 + 4x - 3$ from first principles.
The definition of the derivative from first principles is:
$$\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

STEP 15

Substitute $y = 2x^2 + 4x - 3$ into the definition.
$$f(x+h) = 2(x+h)^2 + 4(x+h) - 3$$

STEP 16

Expand $f(x+h)$.
$$f(x+h) = 2(x^2 + 2xh + h^2) + 4x + 4h - 3$$ $$f(x+h) = 2x^2 + 4xh + 2h^2 + 4x + 4h - 3$$

STEP 17

Subtract $f(x)$ from $f(x+h)$.
$$f(x+h) - f(x) = (2x^2 + 4xh + 2h^2 + 4x + 4h - 3) - (2x^2 + 4x - 3)$$ $$f(x+h) - f(x) = 4xh + 2h^2 + 4h$$

STEP 18

Divide by $h$.
$$\frac{f(x+h) - f(x)}{h} = \frac{4xh + 2h^2 + 4h}{h} = 4x + 2h + 4$$

STEP 19

Take the limit as $h$ approaches 0.
$$\lim_{h \to 0} (4x + 2h + 4) = 4x + 4$$ The derivative of $y = 2x^2 + 4x - 3$ is $\frac{dy}{dx} = 4x + 4$.

STEP 20

For part (c), we need to use integration by substitution to evaluate $\int_{0}^{1} (3x^2 + 2)^4 \, dx$.
Let $u = 3x^2 + 2$.

STEP 21

Differentiate $u$ with respect to $x$.
$$\frac{du}{dx} = 6x$$ $$du = 6x \, dx$$

STEP 22

Rewrite the integral in terms of $u$.
$$\int_{0}^{1} (3x^2 + 2)^4 \, dx = \int_{0}^{1} u^4 \cdot \frac{du}{6x}$$

STEP 23

Since $u = 3x^2 + 2$, when $x = 0$, $u = 2$, and when $x = 1$, $u = 5$.
$$\int_{2}^{5} u^4 \cdot \frac{du}{6x}$$

STEP 24

Since $du = 6x \, dx$, the integral simplifies to:
$$\int_{2}^{5} \frac{u^4}{6} \, du$$

STEP 25

Factor out the constant $\frac{1}{6}$.
$$\frac{1}{6} \int_{2}^{5} u^4 \, du$$

STEP 26

Integrate $u^4$.
$$\frac{1}{6} \left[ \frac{u^5}{5} \right]_{2}^{5}$$

STEP 27

Evaluate the definite integral.
$$\frac{1}{6} \left( \frac{5^5}{5} - \frac{2^5}{5} \right)$$ $$\frac{1}{6} \left( 625 - \frac{32}{5} \right)$$

STEP 28

Simplify the expression.
$$\frac{1}{6} \left( 625 - 6.4 \right)$$ $$\frac{1}{6} \cdot 618.6 = 103.1$$ The value of the integral is $103.1$.

STEP 29

For part (d), we need to use the product rule to find $\frac{dy}{dx}$ for the given functions.
i. $y = (3x^2 + 1)(5t - 8)$
The product rule states:
$$\frac{d}{dx} [u \cdot v] = u' \cdot v + u \cdot v'$$ Let $u = 3x^2 + 1$ and $v = 5t - 8$.

STEP 30

Differentiate $u$ and $v$.
$$u' = 6x$$ $$v' = 5$$

STEP 31

Apply the product rule.
$$\frac{dy}{dx} = (6x)(5t - 8) + (3x^2 + 1)(5)$$ $$\frac{dy}{dx} = 30xt - 48x + 15x^2 + 5$$ ii. $y = t^2 \sin t$
Let $u = t^2$ and $v = \sin t$.

STEP 32

Differentiate $u$ and $v$.
$$u' = 2t$$ $$v' = \cos t$$

Apply the product rule.
$$\frac{dy}{dx} = (2t)(\sin t) + (t^2)(\cos t)$$ $$\frac{dy}{dx} = 2t \sin t + t^2 \cos t$$