Find the intervals on which \( f(x) \) is increasing and the intervals on which \( f(x) \) is decreasing. Then sketch the graph. Add horizontal tangent lines \( f(x)=2 x^{4}-100 x^{2} \) Compute the derivative of \( f(x) \). \( f^{\prime}(x)=\square \)

The first step is to compute the derivative of \[ f(x)=2x^{4}-100x^{2}. \] **Step 1. Compute the derivative** Differentiate each term: - For \(2x^4\), the derivative is \(8x^3\) (since \( \frac{d}{dx}(x^4)=4x^3 \) and \( 2\cdot4=8 \)). - For \(-100x^2\), the derivative is \(-200x\) (since \( \frac{d}{dx}(x^2)=2x \) and \( -100\cdot2=-200 \)). Thus, the derivative is: \[ f'(x)=8x^3-200x. \] **Step 2. Factor the derivative** Factor out the common factor \(8x\): \[ f'(x)=8x^3-200x=8x\left(x^2-25\right). \] Recognize that \(x^2-25\) is a difference of squares: \[ x^2-25=(x-5)(x+5). \] So the fully factored form is: \[ f'(x)=8x(x-5)(x+5). \] **Step 3. Find the critical points** Set \(f'(x)=0\): \[ 8x(x-5)(x+5)=0. \] This gives us: \[ x=0,\quad x-5=0\Rightarrow x=5,\quad x+5=0\Rightarrow x=-5. \] **Step 4. Determine the intervals of increase and decrease** Use test points in the intervals determined by the critical points: - **For \(x < -5\)**, choose \(x=-6\): \[ f'(-6)=8(-6)(-6-5)(-6+5)=8(-6)(-11)(-1). \] Since the product of three negatives is negative, \(f'(-6)<0\). Thus, \(f(x)\) is decreasing on \((-\infty, -5)\). - **For \(-5 < x < 0\)**, choose \(x=-3\): \[ f'(-3)=8(-3)(-3-5)(-3+5)=8(-3)(-8)(2). \] Here, two negatives and one positive yield a positive result, so \(f'(-3)>0\). Thus, \(f(x)\) is increasing on \((-5, 0)\). - **For \(0 < x < 5\)**, choose \(x=3\): \[ f'(3)=8(3)(3-5)(3+5)=8(3)(-2)(8). \] One negative factor gives \(f'(3)<0\). Thus, \(f(x)\) is decreasing on \((0, 5)\). - **For \(x > 5\)**, choose \(x=6\): \[ f'(6)=8(6)(6-5)(6+5)=8(6)(1)(11). \] All factors are positive so \(f'(6)>0\). Thus, \(f(x)\) is increasing on \((5, \infty)\). **Step 5. Identify the horizontal tangent lines** Horizontal tangent lines occur where \(f'(x)=0\), at \(x=-5\), \(x=0\), and \(x=5\). Calculate the corresponding \(y\)-values: - For \(x=-5\): \[ f(-5)=2(-5)^4-100(-5)^2=2(625)-100(25)=1250-2500=-1250. \] - For \(x=0\): \[ f(0)=2(0)^4-100(0)^2=0. \] - For \(x=5\): \[ f(5)=2(5)^4-100(5)^2=2(625)-100(25)=1250-2500=-1250. \] Thus, the horizontal tangent lines are: \[ y=-1250 \quad \text{(at \(x=-5\) and \(x=5\))},\quad y=0 \quad \text{(at \(x=0\))}. \] **Summary** - The derivative is: \[ f'(x)=8x^3-200x. \] - Increasing on the intervals: \((-5, 0)\) and \((5, \infty)\). - Decreasing on the intervals: \((-\infty, -5)\) and \((0, 5)\). - Horizontal tangent lines occur at: \[ x=-5 \; (y=-1250), \quad x=0 \; (y=0), \quad x=5 \; (y=-1250). \] The graph of \(f(x)\) would show: - A local minimum at \(x=-5\) and \(x=5\) (both at \(y=-1250\)). - A local maximum at \(x=0\) (at \(y=0\)). - The curve decreasing before \(x=-5\), increasing between \(-5\) and \(0\), decreasing between \(0\) and \(5\), and increasing after \(x=5\). Thus, \[ f^{\prime}(x)=8x^3-200x. \]