If f(x) is a polynomial such that f(1) =1, f(2) =2, f(3) =3 and f(4) =16. Find the value of f(5)?
Let us assume : [math]f(x) = ax^3 + bx^2 + cx + d [/math]Then we get, [math]f(1) = a + b + c + d = 1 ... (1)[/math][math]f(2) = 8a + 4b + 2c + d = 2 ... (2)[/math][math]f(3) = 27a + 9b + 3c + d = 3 ...(3)[/math][math]f (4) = 64a + 16b + 4c + d = 16 ...(4)[/math]Solving (1) and (2), We get[math]7a + 3b + c = 1 ...(5)[/math]Solving (1) and (3), We get[math]26a + 8b + 2c = 2 ...(6)[/math]Solving (1) and (4), We get[math]63a + 15b + 3c = 15 ...(7)[/math]Solving (5) and (6)[math]6a + b = 0[/math]Solving (5) and (7)[math]7a + b = 2[/math]This gives, a = 2, b = -12And then, c = 23 and d = - 12Thus,[math] f(x) = 2x^3 - 12x^2 + 23x - 12[/math]And So, [math]f(5) = 2*(5)^3 - 12*(5)^2 + 23(5) - 12 = 53[/math]The required answer is f(5) = 53
When a polynomial f(x) is divided by (x-1) and (x-2), it leaves remainder 5 and 7 respectively. What is the remainder when divided by (x-1) (x-2)?
F(x) = (x-1) q(x) + 5 …..1F(x) = (x-2) q'(x) + 7 ……2So F(1) = 5 and F(2) = 7.Now F(x) is divided by (x-1)(x-2) it's is a polynomial of degree two.Reminder should be a polynomial of degree less then 2. Say.. r=ax+bNow.. F(x) = (x-1)(x-2) q”(x) + rF(x) = (x-1)(x-2) q”(x) + (ax+b). ………3@x=1. 5= a+b@x=2. 7= 2a+bSolve two equations and get the value of a and b. That's the ans.r = 2x+3.
Consider f(x)=10−e^x. Find slope and tangent line.?
A. First, take the derivative of f(x). f'(x) = -e^x Find where f(x) crosses the x axis by setting f(x) equal to 0. 10 - e^x = 0 e^x = 10 x = ln 10 So the coordinates are (ln 10, 0) f'(ln 10) is the slope of the tangent line f'(ln 10) = -e^(ln 10) = -10 Slope of tangent line is -10. B. To find the equation of the tangent line use the point slope formula. y - y1 = m(x - x1) y - 0 = -10(x - ln 10) y = -10x - 10*ln 10 C. To find the equation of the line perpendicular to the tangent line, use the negative reciprical of the slope of the tangent line. This slope is 0.1. Use the same point slope formula and same point to find the equation. y - y1 = m(x - x1) y - 0 = 0.1(x - ln 10) y = 0.1x- 0.1*ln 10
If f(1) =1, f(2) =2,f(3) =3 ,f(4) =5, what is f(x)?
f(1)=1f(2)=2f(3)=3We observe a pattern i.e f(x)-x=0f(1)-1=0f(2)-2=0f(3)-3=0This is not possible for f(4)=5.Let us take a new polynomial q(x) which is f(x)-x.q(x)=f(x)-xSo, when x = 4q(4)=f(4)-4q(4)=5–4=1From (1)….q(x)=k(x-1)(x-2)(x-3)-xq(4)=k(3)(2)(1)-45=6kk=5/6From (2)..q(5)+5=k(4)(3)(2)q(5)+5=24*5/6q(5)=15From (1)…q(5)=f(5)-515=f(5)-520=f(5)
If p(x) is the polynomial of degree 4 with leading coefficient as three such that P(1) = 2,p(2) = 8, P(3) = 18 P(4) = 32, then how do you find the value of P(5)?
Let P(x) = 3x^4 + ax^3 + bx^2 + cx + d. Putting x= 1,2,3 and 4 and equating with the given values of P(1), P(2), P(3) & P(4) respectively you will get 4 equations asa+b+c+d = - 1 ..... (1)8a +4b+2c+d = - 40 ...... (2)27a+9b+3c+d = - 225 ...... (3) and64a+16b+4c+d = -736 ..... (4)Solving these equations you may get (please check)a=-17, b=-10, c= 188 and d= -162 Put these values in P(5) i.e. 1875+125a+25b+5c+d to get the answer.
If f(x)-2f(1-x) = x²+2, then what is f(x)?
Given[math]f(x)-2f(1-x)=x^2+2\tag 1[/math]Setting [math]x=1-x[/math] then we get[math]f(1-x)-2f(1-1+x)=(1-x)^2+2[/math][math]f(1-x)-2f(x)=x^2-2x+3[/math][math]2f(1-x)-4f(x)=2x^2-4x+6\tag2[/math]Adding (1) & (2), we get[math]-3f(x)=3x^2-4x+8[/math][math]f(x)=-x^2+\frac{4}{3}x-\frac83[/math]
How can you differentiate log base 10?
[math]\frac{\mathrm{d}}{\mathrm{d}x}(\log_{10}{x})[/math][math]= \frac{\mathrm{d}}{{\mathrm{d}x}}(\frac{\ln{x}}{\ln{10}}) \because \log_a{x} = \frac{\log_b{x}}{\log_b{a}}[/math][math]= \frac{1}{\ln{10}} \frac{\mathrm{d}}{{\mathrm{d}x}}(\ln{x})[/math][math]= \frac{1}{\ln{10}} \cdot \frac{1}{x}[/math][math]= \frac{1}{x\ln{10}}[/math]
What is the number of subsets of S= {1, 2, 3…10} that contain two or three elements?
[math]n(S)=10[/math]Number of subsets of size 2 = Number of ways to choose 2 elements from 10 elements = [math]^{10}C_2[/math]Number of subsets of size 3 = Number of ways to choose 3 elements from 10 elements = [math]^{10}C_3[/math]Number of subsets of size 2 or size 3 = [math]^{10}C_2+^{10}C_3 [/math][math]= \dfrac{10!}{(10-2)!2!} + \dfrac{10!}{(10-3)!3!}[/math][math]\left[ ^nC_r = \dfrac{n!}{(n-r)!r!} \right][/math][math]= \dfrac{10×9×8!}{8!×2!} + \dfrac{10×9×8×7!}{7!×3!}[/math][math]= \dfrac{10×9}{2×1} + \dfrac{10×9×8}{3×2×1}[/math][math]= \dfrac{90}{2} + \dfrac{10×3×4}{1}[/math][math]= 45+120[/math][math]= 165[/math][We use combination because order of elements isn't important. For example, {6,3} and {3,6} denote the same set].
What is the maximum value of (logx)/x in the interval [2,∞)?
I am assuming that the log(x) is a log base 10, rather than a natural log.The derivative of log(x)/x equals log(x)/(-1/x^2)+(1/x)*[1/(x*ln(10)] = (1/x^2) * [1/ln(10)-log(x)].Setting this to zero will yield an x that is either a maximum, a minimum, or a saddle point. We need only consider the quantity within the brackets:[1/ln(10)-log(x)] = 0log(x) = ln(x)/ln(10) = 1/ln(10)Therefore, ln(x)=1 and x = e.The second derivative equals (1/x^2) * [-1/(x*ln(10))] - (2/x^3) *[1/ln(10)-log(x)]. Substituting x = e sets the second term to zero, and the first term is negative. So x = e is a maximum.Showing that it is a maximum over the interval [2, infinity) requires that the first derivative be positive for x < e and negative for x > e.Over these intervals, 1/x^2 is always positive. So the sign of the first derivative depends on the sign of 1/ln(10)-log(x).For x < e, log(x) = ln(x)/ln(10), which is less than 1/ln(10), because ln(x) < 1 — hence, the derivative is positive. For x > e, log(x) > 1/ln(10), and the derivative is negative.