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# Tutor profile: Ali K.

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Ali K.
Mathematics Tutor and Biology Lover
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## Questions

### Subject:Pre-Calculus

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Question:

Find all real values of $$x$$ that satisfy the inequality $$2x^3+3x\leq 3x^2+1$$.

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Ali K.

We may first rewrite the given inequality as $$2x^3-3x^2+3x-1\leq 0$$. Next, we attempt to factor the cubic polynomial $$f(x)=2x^3-3x^2+3x-1$$. Since all the coefficients of our polynomial are integers, we can invoke the rational root theorem: If $$f(x)$$ has any rational roots, they will each be of the form $$\frac{p}{q}$$, where $$p$$ divides the polynomial's constant term, $$-1$$, and $$q$$ divides the leading coefficient, $$2$$. The list of candidate rational roots for our polynomial is therefore $$1$$, $$-1$$, $$\frac{1}{2}$$, $$-\frac{1}{2}$$. Evaluating $$f$$ at these values reveals that $$f(\frac{1}{2})=0$$. Therefore $$\frac{1}{2}$$ is a root of our polynomial and $$2x-1$$ is a factor of our polynomial. Dividing $$2x^3-3x^2+3x-1$$ by $$2x-1$$ (using polynomial long division or synthetic division) yields the quotient $$x^2-x+1$$ (with no remainder). Therefore $$f(x)=(2x-1)(x^2-x+1)$$. Let us consider the quadratic factor, $$q(x)=x^2-x+1$$, on its own. The graph of $$y=q(x)$$ is a parabola, and this parabola opens upward (since the leading coefficient, $$1$$, is positive). Its discriminant, $$(-1)^2-4(1)(1)=-3$$, is negative, so $$q$$ has no real zeros: the graph of $$y=q(x)$$ never intersects the $$x$$ axis. We conclude that the graph of $$y=q(x)$$ lies entirely above the $$x$$ axis. In other words, $$q(x)$$ is positive for all real values of $$x$$. Since $$f(x)=(2x-1)q(x)$$ and $$q(x)$$ is always positive, we see that $$f(x)\leq 0$$ if and only if $$2x-1\leq 0$$. The given inequality is therefore satisfied if and only if $$x\leq\frac{1}{2}$$.

### Subject:Calculus

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Question:

Determine the global (absolute) maximum and minimum values of the function $$f$$ defined by $$f(x)=3x^{2/3}(2x+5),$$ $$-2\leq x\leq 1$$.

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Ali K.

The domain of $$f$$ is the closed interval $$[-2,1]$$. Since $$f$$ is a product of functions that are continuous on that interval, it follows that $$f$$ is continuous on its domain. We may therefore apply the "closed interval method" to locate the global maximum and minimum values of $$f$$. First, we differentiate $$f$$, perhaps by invoking the product rule: $$f^\prime(x)=3\cdot \frac{2}{3}x^{-\frac{1}{3}}(2x+5)+3x^{\frac{2}{3}}(2)=x^{-\frac{1}{3}}(4x+10)+x^{-\frac{1}{3}}(6x)=10x^{-\frac{1}{3}}(x+1)$$, valid for $$-2\leq x < 0$$ and for $$0<x\leq 1$$. $$f^\prime$$ is undefined at $$0$$. (The graph of $$f$$ exhibits a cusp there.) Next, we locate the critical numbers of $$f$$ (where $$f^\prime$$ is zero or undefined): $$f^\prime$$ is zero only at $$-1$$, and it is undefined only at $$0$$. Now, we evaluate $$f$$ at its critical numbers and at the endpoints of its closed interval domain: $$f(-2)=3(-2)^{2/3}[2(-2)+5]=3\sqrt{4}\approx 4.7622$$, $$f(-1)=3(-1)^{2/3}[2(-1)+5]=9$$, $$f(0)=3(0)^{2/3}[2(0)+5]=0$$, $$f(1)=3(1)^{2/3}[2(1)+5]=21$$. Finally, we conclude that the largest of these values, $$21$$, is the global maximum value of $$f$$, while the smallest, $$0$$, is the global minimum value of $$f$$.

### Subject:Algebra

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Question:

Find all real values of $$x$$ that satisfy the equation $$2e^{6x}=3+5e^{3x}$$.

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Ali K.

We observe that $$e^{6x}=(e^{3x})^2$$, so the given equation may be written as $$2(e^{3x})^2=3+5e^{3x}$$. Letting $$z=e^{3x}$$, we have $$2z^2=3+5z$$. Rearranging, we get $$2z^2-5z-3=0$$. Factoring, we have $$(2z+1)(z-3)=0$$. Therefore $$z=-\frac{1}{2}$$ or $$z=3$$. That is, $$e^{3x}=-\frac{1}{2}$$ $$(*)$$ or $$e^{3x}=3$$ $$(**)$$. Since $$e^{3x}$$ is positive for all real $$x$$, we see that $$(*)$$ is inadmissible. Only $$(**)$$ is admissible. From it, we find $$3x=\ln(3)$$ and hence $$x=\frac{1}{3}\ln(3)$$.

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