answer：Okay, so I have this problem here about the graph of the function f(x) = 2x³ - 7. The question is asking how the graph goes, with options from A to E. I need to figure out whether it goes up to the right and down to the left, down to the right and up to the left, up both ways, down both ways, or none of these.Alright, let's start by recalling what the graph of a cubic function generally looks like. I remember that cubic functions have an "S" shape, but depending on the leading coefficient, they can be oriented differently. The leading term is the term with the highest power of x, which in this case is 2x³. The coefficient here is positive, which I think affects the direction the graph goes.So, for a cubic function like f(x) = ax³ + bx² + cx + d, the leading term is ax³. If a is positive, the graph will go up to the right and down to the left. If a is negative, it will go down to the right and up to the left. Let me make sure I remember that correctly. Yeah, I think that's right because as x becomes very large positive, the x³ term dominates, so if a is positive, f(x) becomes very large positive, meaning it goes up to the right. Similarly, as x becomes very large negative, x³ becomes very large negative, so f(x) becomes very large negative, meaning it goes down to the left.In this function, f(x) = 2x³ - 7, the leading coefficient is 2, which is positive. So according to what I just thought, the graph should go up to the right and down to the left. Let me check the options:A) up to the right and down to the leftB) down to the right and up to the leftC) up to the right and up to the leftD) down to the right and down to the leftE) none of these ways.So, based on my reasoning, option A seems correct. But wait, let me make sure I'm not making a mistake. Maybe I should sketch a rough graph or think about specific points.Let's plug in some values. When x is 0, f(0) = 2(0)³ - 7 = -7. So the graph passes through (0, -7). When x is 1, f(1) = 2(1)³ - 7 = 2 - 7 = -5. When x is 2, f(2) = 2(8) - 7 = 16 - 7 = 9. So as x increases, f(x) increases from -7 to -5 to 9, which is going up.When x is -1, f(-1) = 2(-1)³ - 7 = -2 - 7 = -9. When x is -2, f(-2) = 2(-8) - 7 = -16 - 7 = -23. So as x becomes more negative, f(x) decreases, going down.This seems to confirm that as x increases to the right, the function goes up, and as x decreases to the left, the function goes down. So the graph should go up to the right and down to the left, which is option A.Wait, but just to be thorough, let me think about the derivative. The derivative of f(x) is f'(x) = 6x². Since x² is always non-negative, the derivative is always non-negative, meaning the function is always increasing. So, it doesn't have any decreasing intervals, which makes sense for a cubic function with a positive leading coefficient. It starts from the bottom left, increases, and goes to the top right.Therefore, I'm pretty confident that the correct answer is A.

answer：Okay, so I have this problem where I need to find the minimum value of ( x^m + y^m ) given that ( x ) and ( y ) are nonnegative real numbers with a constant sum ( a ). Hmm, let's see. I remember something about inequalities and optimization. Maybe I can use some kind of mean inequality here?First, since ( x + y = a ), I can express ( y ) as ( a - x ). So, the expression becomes ( x^m + (a - x)^m ). I need to find the minimum of this function with respect to ( x ) in the interval ([0, a]).I think calculus might help here. If I take the derivative of ( x^m + (a - x)^m ) with respect to ( x ), set it equal to zero, and solve for ( x ), that should give me the critical points which could be minima or maxima.Let's compute the derivative:[frac{d}{dx} left( x^m + (a - x)^m right) = m x^{m-1} - m (a - x)^{m-1}]Setting this equal to zero:[m x^{m-1} - m (a - x)^{m-1} = 0]Simplify by dividing both sides by ( m ):[x^{m-1} = (a - x)^{m-1}]Taking the ((m-1))-th root of both sides:[x = a - x]Solving for ( x ):[2x = a implies x = frac{a}{2}]So, ( x = frac{a}{2} ) and consequently ( y = frac{a}{2} ). That makes sense; the function is symmetric around ( x = frac{a}{2} ). Now, let's check the second derivative to ensure it's a minimum. The second derivative is:[frac{d^2}{dx^2} left( x^m + (a - x)^m right) = m(m - 1) x^{m-2} + m(m - 1) (a - x)^{m-2}]Since ( m ) is a positive integer, ( m(m - 1) ) is positive for ( m geq 2 ). Also, ( x^{m-2} ) and ( (a - x)^{m-2} ) are nonnegative because ( x ) and ( a - x ) are nonnegative. Therefore, the second derivative is positive, confirming that ( x = frac{a}{2} ) is indeed a point of local minimum.So, substituting ( x = frac{a}{2} ) into the original expression:[left( frac{a}{2} right)^m + left( frac{a}{2} right)^m = 2 left( frac{a}{2} right)^m = frac{a^m}{2^{m-1}}]Alright, that seems to be the minimum value. Let me think if there's another way to approach this, maybe using inequalities like the Power Mean or AM-GM.I recall that for nonnegative numbers, the Power Mean Inequality states that for ( r > s ):[left( frac{x^r + y^r}{2} right)^{1/r} geq left( frac{x^s + y^s}{2} right)^{1/s}]But in this case, we're dealing with the same exponent ( m ) on both terms. Maybe the AM-GM inequality can help here. The AM-GM inequality says that for nonnegative numbers:[frac{x + y}{2} geq sqrt{xy}]But that's for the arithmetic mean and geometric mean. I'm not sure if it directly applies here since we have exponents involved.Wait, maybe using convexity. The function ( f(t) = t^m ) is convex for ( m geq 1 ) because its second derivative is nonnegative. By Jensen's Inequality, for a convex function:[fleft( frac{x + y}{2} right) leq frac{f(x) + f(y)}{2}]But here, we have the sum ( x^m + y^m ). Let me rearrange Jensen's Inequality:[frac{f(x) + f(y)}{2} geq fleft( frac{x + y}{2} right)]Multiplying both sides by 2:[x^m + y^m geq 2 left( frac{x + y}{2} right)^m]Since ( x + y = a ), this becomes:[x^m + y^m geq 2 left( frac{a}{2} right)^m = frac{a^m}{2^{m-1}}]So, this confirms the minimum value we found earlier. Cool, so both calculus and Jensen's Inequality lead us to the same conclusion.Now, moving on to the second problem. It's a generalization of the first one but with more variables. We have ( n ) nonnegative real numbers ( x_1, x_2, ldots, x_n ) with a constant sum ( k ). We need to prove that the minimum of ( x_1^m + x_2^m + cdots + x_n^m ) occurs when all ( x_i ) are equal.Hmm, this seems similar to the first problem but extended to multiple variables. I think the approach would be similar, using either calculus or inequalities.Let me try using the method of Lagrange multipliers since we're dealing with optimization under a constraint. The function to minimize is:[f(x_1, x_2, ldots, x_n) = x_1^m + x_2^m + cdots + x_n^m]Subject to the constraint:[g(x_1, x_2, ldots, x_n) = x_1 + x_2 + cdots + x_n - k = 0]The method of Lagrange multipliers tells us that at the minimum, the gradient of ( f ) is proportional to the gradient of ( g ). So, for each ( i ):[frac{partial f}{partial x_i} = lambda frac{partial g}{partial x_i}]Calculating the partial derivatives:[m x_i^{m-1} = lambda]So, for all ( i ), ( x_i^{m-1} = frac{lambda}{m} ). This implies that all ( x_i ) are equal because their ((m-1))-th powers are equal. Let's denote ( x_i = c ) for all ( i ).Since the sum ( x_1 + x_2 + cdots + x_n = k ), we have:[n c = k implies c = frac{k}{n}]Therefore, each ( x_i = frac{k}{n} ). Substituting back into the function ( f ):[fleft( frac{k}{n}, frac{k}{n}, ldots, frac{k}{n} right) = n left( frac{k}{n} right)^m = frac{k^m}{n^{m-1}}]So, the minimum value is ( frac{k^m}{n^{m-1}} ).Alternatively, using Jensen's Inequality again, since ( f(t) = t^m ) is convex for ( m geq 1 ), we have:[frac{f(x_1) + f(x_2) + cdots + f(x_n)}{n} geq fleft( frac{x_1 + x_2 + cdots + x_n}{n} right)]Multiplying both sides by ( n ):[x_1^m + x_2^m + cdots + x_n^m geq n left( frac{k}{n} right)^m = frac{k^m}{n^{m-1}}]Equality holds when all ( x_i ) are equal, which is when ( x_i = frac{k}{n} ) for all ( i ). So, this confirms the result again.I think both methods are solid. The calculus approach with Lagrange multipliers gives a clear path, and the inequality approach using Jensen's is more elegant but relies on knowing the inequality.Wait, but what if ( m = 1 )? Then, ( x^m + y^m = x + y = a ), so the minimum is just ( a ). In that case, our formula ( frac{a^m}{2^{m-1}} ) gives ( frac{a}{1} = a ), which is correct. Similarly, for ( m = 2 ), the minimum is ( frac{a^2}{2} ), which I know is correct because ( x^2 + y^2 ) is minimized when ( x = y ).Another thought: if ( m = 0 ), but the problem states ( m ) is a positive integer, so we don't have to consider that case. Also, for ( m = 3 ), the minimum would be ( frac{a^3}{4} ), which seems consistent with the pattern.Let me test with specific numbers. Suppose ( a = 2 ) and ( m = 2 ). Then, the minimum should be ( frac{2^2}{2^{2-1}} = frac{4}{2} = 2 ). If ( x = y = 1 ), then ( x^2 + y^2 = 1 + 1 = 2 ), which matches. If I choose ( x = 0 ) and ( y = 2 ), then ( x^2 + y^2 = 0 + 4 = 4 ), which is larger. So, it works.Similarly, for ( a = 4 ), ( m = 3 ). The minimum should be ( frac{4^3}{2^{3-1}} = frac{64}{4} = 16 ). If ( x = y = 2 ), then ( x^3 + y^3 = 8 + 8 = 16 ). If I choose ( x = 0 ) and ( y = 4 ), then ( x^3 + y^3 = 0 + 64 = 64 ), which is way larger. So, it holds.For the second problem, let's take ( n = 3 ), ( k = 3 ), ( m = 2 ). The minimum should be ( frac{3^2}{3^{2-1}} = frac{9}{3} = 3 ). If all ( x_i = 1 ), then the sum is ( 1 + 1 + 1 = 3 ). If I choose ( x_1 = 3 ), ( x_2 = 0 ), ( x_3 = 0 ), then the sum is ( 9 + 0 + 0 = 9 ), which is larger. So, it works here too.I think I've covered the cases, and both problems make sense now. The key idea is that distributing the total sum equally among the variables minimizes the sum of their powers when the exponent is at least 1. This is due to the convexity of the power function, which causes the sum to increase as the variables become more unequal.So, in summary, for both problems, the minimum occurs when all variables are equal, and the minimum value can be calculated using the formulas derived above.**Final Answer**1) The minimum value is boxed{dfrac{a^m}{2^{m-1}}}.2) The minimum value is boxed{dfrac{k^m}{n^{m-1}}}.

answer：Okay, so I have this problem here about lines and points. Let me try to understand it step by step. We have two points, P₁ (x₁, y₁) which is on line l, and P₂ (x₂, y₂) which is not on line l. The equation of line l is given by f(x, y) = 0. Then, there's this equation: f(x, y) - f(x₁, y₁) - f(x₂, y₂) = 0. I need to figure out what this equation represents. The options are:A. A line coinciding with lB. A line parallel to l and passing through P₁C. A line parallel to l and passing through P₂D. A line passing through both points P₁ and P₂Alright, let's break this down. First, since P₁ is on line l, substituting P₁ into the equation f(x, y) = 0 should satisfy it. That means f(x₁, y₁) = 0. So, in the given equation, f(x, y) - f(x₁, y₁) - f(x₂, y₂) = 0, the term f(x₁, y₁) is zero. So, the equation simplifies to f(x, y) - f(x₂, y₂) = 0.Now, f(x, y) = 0 is the equation of line l. If I subtract f(x₂, y₂) from both sides, I get f(x, y) = f(x₂, y₂). Since f(x, y) is the equation of line l, subtracting a constant (f(x₂, y₂)) from it should give me another line that's parallel to l. Because when you have two lines with the same coefficients for x and y but different constants, they are parallel.So, the equation f(x, y) = f(x₂, y₂) is a line parallel to l. Now, does this line pass through P₁ or P₂? Let's check.If I plug in P₁ (x₁, y₁) into the equation f(x, y) = f(x₂, y₂), we get f(x₁, y₁) = f(x₂, y₂). But we know f(x₁, y₁) = 0, so 0 = f(x₂, y₂). But wait, P₂ is not on line l, so f(x₂, y₂) ≠ 0. That means P₁ does not lie on this new line.What about P₂? If I plug in P₂ (x₂, y₂) into f(x, y) = f(x₂, y₂), we get f(x₂, y₂) = f(x₂, y₂), which is true. So, P₂ lies on this new line.Therefore, the equation f(x, y) - f(x₁, y₁) - f(x₂, y₂) = 0 represents a line parallel to l and passing through P₂. So, the correct answer should be option C.

answer：Alright, so I have this inequality to prove: (ax + by)(bx + ay) ≥ xy, where a, b, x, and y are all positive numbers, and a + b = 1. Hmm, okay, let's see how I can approach this.First off, since a and b are positive and add up to 1, they're like weights or something. Maybe I can use that fact somehow. The expression (ax + by)(bx + ay) looks a bit complicated, but maybe if I expand it, things will become clearer.Let me try expanding it step by step. So, (ax + by)(bx + ay) would be ax * bx + ax * ay + by * bx + by * ay. Let's compute each term:1. ax * bx = abx²2. ax * ay = a²xy3. by * bx = b²xy4. by * ay = aby²So, putting it all together, the expanded form is abx² + a²xy + b²xy + aby². Hmm, I can factor out some terms here. Let's see:abx² + aby² = ab(x² + y²)a²xy + b²xy = xy(a² + b²)So, the entire expression becomes ab(x² + y²) + xy(a² + b²). Okay, that's a bit simpler. Now, I need to show that this is greater than or equal to xy.So, ab(x² + y²) + xy(a² + b²) ≥ xy.Hmm, maybe I can rearrange this inequality to make it easier to handle. Let's subtract xy from both sides:ab(x² + y²) + xy(a² + b²) - xy ≥ 0Factor out xy from the last two terms:ab(x² + y²) + xy(a² + b² - 1) ≥ 0Wait, since a + b = 1, maybe I can use that to simplify a² + b² - 1. Let's recall that (a + b)² = a² + 2ab + b². Since a + b = 1, this becomes 1 = a² + 2ab + b². Therefore, a² + b² = 1 - 2ab.So, substituting back into the inequality:ab(x² + y²) + xy(1 - 2ab - 1) ≥ 0Simplify the terms inside the parentheses:ab(x² + y²) + xy(-2ab) ≥ 0Factor out ab:ab(x² + y² - 2xy) ≥ 0Hmm, x² + y² - 2xy is a familiar expression. That's equal to (x - y)², right? So, we have:ab(x - y)² ≥ 0Since a and b are positive, ab is positive. And (x - y)² is always non-negative because any real number squared is non-negative. So, the product of two non-negative numbers is non-negative. Therefore, ab(x - y)² is always greater than or equal to zero.Thus, we've shown that ab(x - y)² ≥ 0, which implies that the original expression (ax + by)(bx + ay) is indeed greater than or equal to xy.Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from the expansion, I got ab(x² + y²) + xy(a² + b²). Then I subtracted xy and factored out xy, leading me to ab(x² + y²) + xy(a² + b² - 1). Using a + b = 1, I found that a² + b² = 1 - 2ab, which seems correct because (a + b)² = a² + 2ab + b² = 1.Substituting that back in, I got ab(x² + y²) - 2abxy, which factors to ab(x² + y² - 2xy) = ab(x - y)². Since both ab and (x - y)² are non-negative, their product is non-negative, so the inequality holds.I think that makes sense. Maybe there's another way to approach it, like using the AM-GM inequality or Cauchy-Schwarz, but this expansion method worked out fine. I don't see any errors in my reasoning, so I feel confident about this proof.

answer：First, I need to determine how many pens Teacher Zhang could buy before the discount. Let's denote the original number of pens he could buy as ( x ).Since the pens are discounted by 25%, the new price is 75% of the original price. This means that with the same amount of money, Teacher Zhang can buy more pens.Given that he can buy 25 more pens with the discounted price, I can set up the equation:[x = 0.75(x + 25)]Expanding and simplifying this equation:[x = 0.75x + 18.75]Subtracting ( 0.75x ) from both sides:[0.25x = 18.75]Finally, solving for ( x ):[x = frac{18.75}{0.25} = 75]Therefore, Teacher Zhang could originally buy 75 pens with the money he brought.

answer：Okay, so I have this geometry problem here about planes and lines. It says that α and β are two different planes, and there's a line l that's in plane α. The question is asking about the condition for line l to be perpendicular to plane β. Specifically, it's asking whether "α perpendicular to β" is a sufficient condition, a necessary condition, both, or neither for "l perpendicular to β."Hmm, let me try to visualize this. I know that two planes being perpendicular means that they intersect at a right angle. So if α is perpendicular to β, that means they form a dihedral angle of 90 degrees. Now, line l is in plane α. So, if line l is perpendicular to plane β, that would mean that line l is perpendicular to every line in plane β that it intersects, right?Wait, but does that necessarily mean that the entire plane α is perpendicular to β? I think there's a theorem about this. Let me recall. I believe that if a line in one plane is perpendicular to another plane, then the two planes are perpendicular. So, if l is in α and l is perpendicular to β, then α must be perpendicular to β. That seems like a sufficient condition because it guarantees that α is perpendicular to β.But is it a necessary condition? That is, does α being perpendicular to β necessarily mean that any line l in α is perpendicular to β? I don't think so. Because if α is perpendicular to β, a line in α could be perpendicular to β, but it could also just be at some other angle or even parallel. It depends on the specific line.So, if α is perpendicular to β, then it's possible for a line in α to be perpendicular to β, but it's not guaranteed. Therefore, α being perpendicular to β is a necessary condition for l to be perpendicular to β, but it's not sufficient on its own because not every line in α will be perpendicular to β.Wait, let me make sure I'm not mixing things up. If l is perpendicular to β, then α must be perpendicular to β. So, in that case, α being perpendicular is a necessary condition for l to be perpendicular. But just because α is perpendicular to β doesn't mean every line in α is perpendicular to β. So, α being perpendicular is necessary but not sufficient.Therefore, the condition "α perpendicular to β" is a necessary but not sufficient condition for "l perpendicular to β." So, the answer should be option B.But just to double-check, let's think about it another way. If I have two perpendicular planes, α and β, then any line in α that is also perpendicular to the line of intersection of α and β would be perpendicular to β. But there are infinitely many lines in α, and only those specific ones would be perpendicular to β. So, it's not that all lines in α are perpendicular to β, just some of them. Therefore, knowing that α is perpendicular to β doesn't guarantee that a particular line l in α is perpendicular to β. It's necessary because if l is perpendicular to β, then α must be, but it's not sufficient because not every line in α will be.Yeah, I think that makes sense. So, the condition is necessary but not sufficient. So, the answer is B.