1. **Problem Statement:** Determine the average gradient of the function $f(x) = x^2 + 2$ between $x=2$ and $x=4$. 2. **Recall the formula for average gradient:** $$\text{Average gradient} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$$ 3. **Calculate $f(4)$ and $f(2)$:** $$f(4) = 4^2 + 2 = 16 + 2 = 18$$ $$f(2) = 2^2 + 2 = 4 + 2 = 6$$ 4. **Substitute values into the average gradient formula:** $$\frac{18 - 6}{4 - 2} = \frac{12}{2} = 6$$ 5. **Final answer:** The average gradient between $x=2$ and $x=4$ is $6$. --- 1. **Problem Statement:** Determine the gradient of $f(x) = 2x^2 - 1$ at the point where $x=1$ using first principles. 2. **Recall the definition of gradient at a point:** $$\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$ 3. **Substitute $a=1$ and $f(x)$:** $$\lim_{h \to 0} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0} \frac{2(1+h)^2 - 1 - (2(1)^2 - 1)}{h}$$ 4. **Expand and simplify numerator:** $$(1+h)^2 = 1 + 2h + h^2$$ $$2(1+h)^2 - 1 = 2(1 + 2h + h^2) - 1 = 2 + 4h + 2h^2 - 1 = 1 + 4h + 2h^2$$ $$2(1)^2 - 1 = 2 - 1 = 1$$ 5. **Substitute back:** $$\lim_{h \to 0} \frac{(1 + 4h + 2h^2) - 1}{h} = \lim_{h \to 0} \frac{4h + 2h^2}{h}$$ 6. **Factor and simplify:** $$\lim_{h \to 0} \frac{h(4 + 2h)}{h} = \lim_{h \to 0} (4 + 2h)$$ 7. **Evaluate the limit:** $$4 + 2(0) = 4$$ 8. **Final answer:** The gradient of $f(x) = 2x^2 - 1$ at $x=1$ is $4$. --- 1. **Problem Statement:** Find the derivative of $f(x) = x^2 + 4$ using first principles. 2. **Recall the definition of derivative:** $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ 3. **Calculate $f(x+h)$:** $$f(x+h) = (x+h)^2 + 4 = x^2 + 2xh + h^2 + 4$$ 4. **Substitute into the limit:** $$f'(x) = \lim_{h \to 0} \frac{(x^2 + 2xh + h^2 + 4) - (x^2 + 4)}{h} = \lim_{h \to 0} \frac{2xh + h^2}{h}$$ 5. **Factor and simplify:** $$\lim_{h \to 0} \frac{h(2x + h)}{h} = \lim_{h \to 0} (2x + h)$$ 6. **Evaluate the limit:** $$2x + 0 = 2x$$ 7. **Final answer:** The derivative of $f(x) = x^2 + 4$ is $f'(x) = 2x$. --- 1. **Problem Statement:** Find the derivative of $f(x) = \frac{1}{x}$ using first principles. 2. **Recall the definition of derivative:** $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ 3. **Calculate $f(x+h)$:** $$f(x+h) = \frac{1}{x+h}$$ 4. **Substitute into the limit:** $$f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h}$$ 5. **Find common denominator and simplify numerator:** $$\frac{1}{x+h} - \frac{1}{x} = \frac{x - (x+h)}{x(x+h)} = \frac{-h}{x(x+h)}$$ 6. **Rewrite the limit:** $$f'(x) = \lim_{h \to 0} \frac{-h}{x(x+h)} \times \frac{1}{h} = \lim_{h \to 0} \frac{-1}{x(x+h)}$$ 7. **Evaluate the limit:** $$-\frac{1}{x(x+0)} = -\frac{1}{x^2}$$ 8. **Final answer:** The derivative of $f(x) = \frac{1}{x}$ is $f'(x) = -\frac{1}{x^2}$. --- **Summary:** - Average gradient between two points is calculated by difference in $y$ over difference in $x$. - Gradient at a point is found using the limit definition of derivative. - Derivatives can be found using first principles by evaluating the limit of the difference quotient. - The derivative of $x^2 + 4$ is $2x$. - The derivative of $\frac{1}{x}$ is $-\frac{1}{x^2}$.