1. **State the problem:** We have two functions defined by integrals: $$g(x) = \int_1^x (5 - 8\sqrt{\ln t}) \, dt, \quad x > 1$$ $$h(x) = \int_1^{x^2} (5 - 8\sqrt{\ln t}) \, dt, \quad x > 1$$ We need to: (a) Find the equation of the tangent line to $g$ at $x=3$. (b) Find $h'(x)$. (c) Determine where $g$ is decreasing. (d) Find relative extrema of $h$ and classify them. --- 2. **Recall formulas and rules:** - By the Fundamental Theorem of Calculus, if $G(x) = \int_a^x f(t) dt$, then $G'(x) = f(x)$. - For a function defined as $H(x) = \int_a^{u(x)} f(t) dt$, by the chain rule: $$H'(x) = f(u(x)) \cdot u'(x)$$ - The equation of the tangent line to $g$ at $x=x_1$ is: $$y - g(x_1) = g'(x_1)(x - x_1)$$ - A function is decreasing where its derivative is negative. - Relative extrema occur where the derivative is zero and the sign of the derivative changes. --- 3. **Part (a): Equation of tangent to $g$ at $x=3$** - Compute $g'(x)$: $$g'(x) = 5 - 8\sqrt{\ln x}$$ - Evaluate $g'(3)$: $$g'(3) = 5 - 8\sqrt{\ln 3}$$ - Compute $g(3)$: $$g(3) = \int_1^3 (5 - 8\sqrt{\ln t}) dt$$ (This integral is left unevaluated as no closed form is requested.) - Tangent line equation: $$y - g(3) = (5 - 8\sqrt{\ln 3})(x - 3)$$ --- 4. **Part (b): Find $h'(x)$** - Since $h(x) = \int_1^{x^2} (5 - 8\sqrt{\ln t}) dt$, by chain rule: $$h'(x) = (5 - 8\sqrt{\ln (x^2)}) \cdot \frac{d}{dx}(x^2) = (5 - 8\sqrt{\ln (x^2)}) \cdot 2x$$ - Simplify inside the square root: $$\ln (x^2) = 2 \ln x$$ - So: $$h'(x) = 2x \left(5 - 8\sqrt{2 \ln x}\right)$$ --- 5. **Part (c): Intervals where $g$ is decreasing** - $g$ is decreasing where $g'(x) < 0$: $$5 - 8\sqrt{\ln x} < 0$$ $$8\sqrt{\ln x} > 5$$ $$\sqrt{\ln x} > \frac{5}{8}$$ $$\ln x > \left(\frac{5}{8}\right)^2 = \frac{25}{64}$$ $$x > e^{25/64}$$ - Since $x > 1$, the open interval where $g$ is decreasing is: $$(e^{25/64}, \infty)$$ --- 6. **Part (d): Relative extrema of $h$** - Find critical points where $h'(x) = 0$: $$2x \left(5 - 8\sqrt{2 \ln x}\right) = 0$$ - Since $x > 1$, $2x \neq 0$, so: $$5 - 8\sqrt{2 \ln x} = 0$$ $$8\sqrt{2 \ln x} = 5$$ $$\sqrt{2 \ln x} = \frac{5}{8}$$ $$2 \ln x = \left(\frac{5}{8}\right)^2 = \frac{25}{64}$$ $$\ln x = \frac{25}{128}$$ $$x = e^{25/128}$$ - To classify extrema, check sign of $h'(x)$ around $x = e^{25/128}$: - For $x < e^{25/128}$, $5 - 8\sqrt{2 \ln x} > 0$ so $h'(x) > 0$. - For $x > e^{25/128}$, $5 - 8\sqrt{2 \ln x} < 0$ so $h'(x) < 0$. - Derivative changes from positive to negative, so $h$ has a **relative maximum** at: $$x = e^{25/128}$$ --- **Final answers:** (a) $$y - g(3) = (5 - 8\sqrt{\ln 3})(x - 3)$$ (b) $$h'(x) = 2x \left(5 - 8\sqrt{2 \ln x}\right)$$ (c) $$g \text{ is decreasing on } (e^{25/64}, \infty)$$ (d) $$h \text{ has a relative maximum at } x = e^{25/128}$$