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MathSimply plug the function into a graphing tool, and you should get this graph.

MathSince you already have the same bases, you just need to set the exponents equal to each other. `3x+2=3 ` Solve for `x ` `x=1/3 `

MathSince you already have the same bases, you just need to set the exponents eqal to each other. `2x1=4 ` Solve for `x ` `x=5/2 `

MathSince you already have the same bases, you just need to set the exponents eqal to each other. `x^(2)3=2x ` Subtract `2x ` from both sides `x^22x3=0 ` Solve the quadratic equation `(x3)(x+1)=0...

MathWe have `e^f(x)=e^g(x) ` which means that `f(x)=g(x) ` , where `f(x)=x^2+6 ` and `g(x)=5x `. To find the value of `x ` such that `x^2+6=5x ` we may plot the difference `x^25x+6=0 `: It is easy...

MathPlug in the value `x=1.4 ` for the function `f(x) ` `f(x)=0.9^(1.4) ` Simplify `f(x)=0.863 `

MathPlug in the value `x=3/2 ` for the function `f(x) ` `f(x)=2.3^(3/2) ` Simplify `f(x)=3.488 `

MathPlug in the value `x=pi ` for the function `f(x) ` `f(x)=5^(pi) ` Simplify `f(x)=0.006 `

MathPlug in the value `x=3/10 ` for the function ` f(x)` `f(x)=(2/3)^(5*(3/10)) ` Simplify ` f(x)=0.544`

MathPlug in the value `x=1.5 ` in for the function `g(x) ` `g(x)=5000(2^(1.5)) ` Simplify `g(x)=1767.767 `

MathPlug in the value `x=24 ` for the function `f(x) ` `f(x)=200(1.2)^(12*24) ` Simplify `f(x)=1.274 E25 `

Math`3^(x+1)=27 ` Get the same base ` 3^(x+1)=3^3` Set the exponents equal to each other `x+1=3 ` Solve for x `x=4 `

Math`2^(x3)=16 ` Get the same base `2^(x3)=2^4 ` Set the exponents equal to each other `x3=4 ` Solve for x `x=7 `

MathGet the bases the same 1. `(1/2^(x))=32 ` Bring up the `2^x ` as a `2^(x) ` 2. `2^(x)=32 ` Get 32 into the base of 2 `2^(x)=2^5 ` Set the exponents equal to each other `x=5 ` Solve for x and do...

Math`5^(x2)=1/125 ` Get same base `5^(x2)=5^(3) ` Set the exponents equal to each other ` x2=3` Solve for x x=1

MathYou can see from the graphs that g(x) has been shifted up 1 unit.

MathYou can see from the graphs that g(x) has been reflected over the yaxis and shifted 3 units.

MathYou can see from the graphs that it has been flipped over the xaxis.

MathAs you can see from the graphs, it is clear that `` is flipped over the xaxis and shifted up 5 units.

MathPlug in `x=3.2 ` in for ` x` in the function. `f(x)=e^(3.2) ` Simplify, ` f(x)=24.532`

MathPlug in the value ` x=240` in for ` x` in the function. `f(x)=1.5e^(240/2) ` Simplify, `f(x)=1.956 E52`

MathPlug in the value `x=6 ` in for ` x` in the function `f(x)=5000e^(0.06*6) ` Simplify, `f(x)=7166.647 `

MathPlug in the value ` x=20` for ` x` in the function `f(x)=250e^(0.05*20) ` Simplify, ` f(x)=679.570`

Matha) To calculate the height in a vertical motion, we apply the following formula: h = (v^2 – v0^2)/2g (1) where: v → is the velocity at any instant of movement. At the highest point, the...

MathThe given line is : 2x  y + 1 = 0 or, y = 2x + 1 (the line is represented in slope intercept form) Thus, the slope of the line = 2 Now, the tangent to the curve f(x) = (x^2) is parallel to the...

MathThe given line is : 4x + y + 3 = 0 or, y = 4x  3 (the line is represented in slope intercept form) Thus, the slope of the line = 4 Now, the tangent to the curve f(x) = 2(x^2) is parallel to...

MathThe given line is : 3x  y + 1 = 0 or, y = 3x + 1 (the line is represented in slope intercept form) Thus, the slope of the line = 3 Now, the tangent to the curve f(x) = (x^3) is parallel to the...

MathThe slope of the line 3xy4=0 and the lines parallel to it will be same. 3xy4=0 y+4=3x y=3x4 Therefore the slope(m) of the line is the coefficient of the x = 3 The slope of the tangent line to...

MathThe given line is : x + 2y  6 = 0 or, y = (1/2)x + 3 (the line is represented in slope intercept form) Thus, the slope of the line = (1/2) Now, the tangent to the curve f(x) = 1/(x^(1/2)) is...

MathSince the tangent line to function f(x) = 1 / sqrt(x 1) has to be parallel to the line x + 2y + 7 = 0, both needs to have the same slope. Equation of the straight line can be rewritten in...

MathUsing the alternative definition of the derivative, the approximation of the derivative given by `lim_(x>3)` `(f(x) f(c))/(xc)` = `lim_(x>3)(x^2 5  (3^2 5))/(x3)` ` ` `=(x^2 5...

Math`lim_(x>1)(f(x)f(c))/(xc)` `lim_(x>1)(x^2  x)/(x  1)` ``Simplify: `lim_(x>1)(x(x1))/(x1)` `lim_(x>1)(x) = 1`

Math`lim_(x>2) (f(x)  f(c))/(xc)` ` ` `lim_(x>2) (x^3 + 2x^2)/(x+2)` Simplify using synthetic division.` ` `lim_(x>2) x^2 = 4`

Math`lim_(x>2) (f(x)  f(c))/(xc)` `lim_(x>2) (x^3 + 6x  20)/(x2)` Simplify using synthetic division. `lim_(x> 2) x^2 + 2x + 10 = 18`

Math`lim_(x>0) (g(x)g(c))/(xc)` `lim_(x>0) (sqrt(x))/x` ` ` In order to evaluate the above function, both the left hand limit and right hand limit need to be evaluated. `L.H.L. =...

Math`lim_(x>4)(f(x)  f(c))/(xc)` `lim_(x>4)(3/x  3/4)/(x4)` ` `Simplify expression: `lim_(x>4)((123x)/(4x))/(x4)` `lim_(x>4)(3(4x))/(4x(x4))` `lim_(x>4)((3/(4x))) = 3/16`

Math`lim_(x>6) (f(x)  f(c))/(xc)` `lim_(x>6) ((x6)^(2/3))/(x6)` ` ` Using division of exponents, it simplifies to: `lim_(x>6) (x6)^(1/3) = 0`

Math`lim_(x>3) (g(x)  g(c))/(xc)` `lim_(x>3) ((x+3)^(1/3))/(x+3)` Using division of exponents, it simplifies to: `lim_(x>3) (x+3)^(2/3) = 0` `` ` `

Math`lim_(x>7) (h(x)  h(c))/(xc)` ` ` `lim_(x>7) (x+7)/(x+7)` In order to evaluate the above function, both the left hand limit and right hand limit need to be evaluated. `L.H.L. =...

Math`lim_(x>6) (f(x)  f(c))/(xc)` ` ` `lim_(x>6) (x6)/(x6)` In order to evaluate the above function, both the left hand limit and right hand limit need to be evaluated. `L.H.L. =...

MathBy limit process, the derivative of a function f(x) is : f'(x) = lim h > 0 [{f(x+h)  f(x)}/h] Now, the given function is : h(s) = 3 + (3/2)s Thus, h'(s) = lim t > 0 [{h(s+t) ...

MathBy limit process, the derivative of a function f(x) is : f'(x) = lim h > 0 [{f(x+h)  f(x)}/h] Now, the given function is : f(x) = 5  (3/2)x Thus, f'(x) = lim t > 0 [{f(x+t) ...

MathBy limit process, the derivative of a function f(x) is : f'(x) = lim h > 0 [{f(x+h)  f(x)}/h] Now, the given function is : f(x) = (x^2) + x  3 thus, f'(x) = lim h > 0 [{{(x+h)^2} +...

MathTo find the derivative of the function `f(x)=x^25` ` ` using the limit process, use the formula `m=lim_(h>0)(f(x+h)f(x))/h` ` ` `m=lim_(h>0)(((x+h)^25)(x^25))/(h)`...

MathBy limit process, the derivative of a function f(x) is : f'(x) = lim h > 0 [{f(x+h)  f(x)}/h] Now, the given function is : f(x) = (x^3)  12x thus, f'(x) = lim h > 0 [{{(x+h)^3} ...

MathBy limit process, the derivative of a function f(x) is : f'(x) = lim h > 0 [{f(x+h)  f(x)}/h] Now, the given function is : f(x) = (x^3) + (x^2) thus, f'(x) = lim h > 0 [{(x+h)^3} +...

Maththe alternate method to find the derivative of the function is the limit form. By limit process, the derivative of a function f(x) is : f'(x) = lim h > 0 [{f(x+h)  f(x)}/h] Now, the given...

MathBy limit process, the derivative of a function f(x) is : `f'(x) = lim_(h > 0) [{f(x+h)  f(x)}/h]` Now, the given function is : `f(x) = 1/(x^2)` thus, `f'(x) = lim_(h > 0) [{{1/(x+h)^2} ...

MathBy limit process, the derivative of a function f(x) is : f'(x) = lim h > 0 [{f(x+h)  f(x)}/h] Now, the given function is : f(x) = sqrt(x+4) THus, f'(x) = lim h > 0 [{f(x+h)  f(x)}/h]...

MathBy limit process, the derivative of a function f(x) is : `f'(x) = lim_(h > 0) [{f(x+h)  f(x)}/h]` Now, the given function is : `f(x) = 4/sqrt(4)` Thus, `f'(x) =lim_(h > 0) [{f(x+h) ...